The Higgs field is a scalar field, $h$, so as far as the Lorentz symmetry goes, it is allowed and expect to interact with any other field. Whatever is the Lorentz-invariant term in the Lagrangian for other fields may be multiplied by $h$ to get an allowed interaction term, often a renormalizable one.
In particular, the Higgs field interacts with all the fermionic fields via the so-called Yukawa interaction, schematically $y\cdot h\bar\psi \psi$, where $\psi$ is a fermion field. The (classically) dimensionless Yukawa couplings $y$ may be large or small. The top quark Yukawa coupling is very large, of order one, which makes the top quark heavy. On the contrary, the Yukawa couplings with the neutrinos is (almost) zero which keeps the neutrinos massless.
In the previous paragraph, I ignored the fact that the Higgs is really a component of the Higgs doublet and the interaction terms has to be gauge-invariant. So a doublet must be contracted with another doublet, and so on. The vev produces masses for one component of the doublet only. There are interactions with $h$ (doublet) and its complex conjugate that make all quarks massive. However, it's not possible to write down the simplest interaction that would give the neutrinos masses in the simplest model. In the simplest model, the neutrinos are left-handed and it's exactly the left-handed part of the neutrinos that can't get the masses. But the neutrinos are still interacting with the Higgs field.
The gauge invariance in fact dictates that the Higgs field has to interact with other fields, the gauge fields. Because the Higgs field carries no color charges like quarks, it's neutral under $SU(3)_{QCD}$ which means that it doesn't have any interactions with gluons.
However, the Higgs field has nonzero charges under the electroweak $SU(2)_W$ because the Higgs field is a doublet; and under the hypercharge $U(1)_Y$ part of the electroweak gauge symmetry. These charges of the Higgs fields mean that the derivatives in its kinetic term have to be replaced by the covariant derivatives and this produces interactions with the electroweak gauge fields.
In general, that makes the gauge bosons for $SU(2)\times U(1)$ massive, too, when the Higgs gets a vacuum expectation value. However, the vev is such that the classical vacuum configuration is invariant under the electromagnetic $U(1)_{em}$ generated by the electric charge $Q = (1/2)Y+T_3$. So under this particular generator, the component of the Higgs field that has a nonzero vev is neutral which is why the Higgs field doesn't directly interact with the corresponding gauge field, the electromagnetic field, and that's why the photon stays massless and the electromagnetic force remains a long-range force. The other three generators of the electroweak group produce gauge bosons $W^\pm, Z^0$ if an orthonormal basis is chosen which is why these three gauge bosons are massive due to the Higgs mechanism.
Here is a free body diagram of the balls:
… and one of the water volume:
The four balance equations are
$$ \begin{align}
B_1 - T_1 - m_1 g & =0 \\
B_2 + T_2 - m_2 g & = 0 \\
F_1 + T_1 - B_1 - M g & = 0 \\
F_2 - B_2 - M g & = 0
\end{align} $$
where $\color{magenta}{B_1}$,$\color{magenta}{B_2}$ are the buoyancy forces, $\color{red}{T_1}$,$\color{red}{T_2}$ are the cord tensions and $M g$ is the weight of the water, $m_1 g$ the weight of the ping pong ball and $m_2 g$ the weight of the steel ball.
Solving the above gives
$$\begin{align}
F_1 & = (M+m_1) g \\
F_2 & = M g + B_2 \\
T_1 & = B_1 - m_1 g \\
T_2 & = m_2 g - B_2
\end{align} $$
So it will tip to the right if the buoyancy of the steel ball $B_2$ is more than the weight of the ping pong ball $m_1 g$.
$$\boxed{F_2-F_1 = B_2 - m_1 g > 0}$$
This is the same answer as @rodrigo but with diagrams and equations.
Best Answer
The usual pop-sci explanation of "the Higgs field exerting a drag force on particles that move through it, sapping their kinetic energy" is unfortunately not very accurate. In technical terms, the mass generation for the weak gauge bosons is really due to the spontaneous symmetry breaking in the Higgs mechanism, and the associated Yukawa couplings generate mass terms for the fermions of the Standard Model (except neutrinos). There is additionally an important distinction to be made between the "Higgs boson" and the "Higgs field", though these are often conflated and even fused in pop-sci descriptions.
I will attempt to explain this in a manner that is more accurate than the pop-sci description while still being accessible. The punchline is:
Although the massive particles continuously interact with the Higgs boson which is present throughout space, this does not give them mass. The Higgs mechanism gives them mass once and for all at the electroweak transition scale.
"Spontaneous symmetry breaking at the electroweak transition scale" roughly means this: at very high energies, the picture of particle physics and phenomenology is very different from ordinary energies. "Very high energies" is entirely equivalent to "very small distances", and we call the energy at which we are looking at the theory the "energy scale".
If, starting from very high energies, you begin to decrease the energy scale, when you reach the electroweak scale at 160 GeV, the Higgs field "condenses". An almost perfect analogy for this phenomenon is how, starting at say 20 °C, you can decrease the temperature of water until it suddenly freezes into ice at 0 °C. Now 160 GeV roughly corresponds to a temperature of $1.85\times 10^{15}\ \mathrm K$, which is really, really high, but still within the domain of the Large Hadron Collider. At this point, and this point alone, the gauge fields and fermion fields gain a mass term, and hence their quanta - the gauge bosons and fundamental fermions - are no longer massless and become massive.
At energies below this scale, they do not have to interact with anything (no pop-sci-esque "Higgs boson fluid") to keep this mass. Indeed, massless and massive particles are fundamentally different in quantum field theory, so a "continuous" interaction type of approach to give mass to particles is doomed right from the get go.
To make sure that e.g. photons do not gain a mass in this process, we encode the exact way that the symmetry breaking takes place into the model.
This choice of symmetry breaking is not arbitrary, nor is it predicted by theory. The whole idea of electroweak spontaneous symmetry breaking is to explain the masses of the $W^\pm$ and $Z$ gauge bosons (and fermions), all with the underlying assumption that the photon is massless, and this is what we put into our model by hand. We could very well have built a similar model where the photon is massive - but we don't, because we don't observe such a thing. Everything need not interact with everything else - the electromagnetic field too permeates throughout spacetime, but does not interact with uncharged particles.
So the real reason that the photon does not gain a mass by the Higgs mechanism is because we don't want it to, otherwise our model would be inaccurate.
Nevertheless, if you are interested in probing deeper, here are
The specifics (but not too technical)
For now, focus only on the nature of the $W^\pm$, $Z$ and $\gamma$ (photon). Ignore the gluons, since they play a mere spectator role in the mass generation - they are unaffected (we say that "the $\mathrm{SU}(3)$ color group remains unbroken"). This is due to the nature of our model building - we should be able to declare that certain particles are massless, and dictate which particles should interact with each other, provided that the resulting model is consistent with the theoretical framework.
Unfortunately, the simplest model that we can build to accommodate the massive nature of the weak force bosons and fermions fails immediately for a straightforward reason: the existence of a mass term violates the gauge symmetry of the fundamental forces, and so our theory is mathematically inconsistent. The Higgs mechanism is the simplest (and only viable) method to explain this, but curiously the photon is not merely a spectator like the gluons and is present in the mechanism.
Prior to spontaneous symmetry breaking, i.e. at energies higher than the electroweak scale, there are four massless gauge fields living peacefully - call them the $W^1, W^2, W^3$ and $B$. In other words, there are four electromagnetic-like forces above 160 GeV, and associated to each one is its own "charge" determining the strength and nature of interactions, exactly the way the photon/electromagnetic field at ordinary energies couples to the electric charge. These gauge fields coexist with the Higgs field - which, most importantly, has a non-zero charge under all of these fields.
At the electroweak transition scale, the Higgs field undergoes spontaneous symmetry breaking. We say that the $\mathrm{SU}(2)_L\times \mathrm U(1)_Y$ symmetry is broken down to $\mathrm U(1)_\mathrm{EM}$. What this means, roughly, is that the original four degrees of freedom (DOF) of the Higgs field break up into a vacuum expectation value with 0 DOF, a Higgs boson with 1 DOF and 3 "Goldstone bosons" with 1 DOF each. This is roughly like how water loses its freedom to flow in different directions after the phase transition, with the DOF "freezing out".
So why don't we observe these Goldstone bosons as physical particles at regular energy scales? Well, owing to the nature of the symmetry breaking, three linear combinations of the four originally massless gauge fields "eat" one Goldstone boson each and become massive, forming the $W^+, W^-$ and $Z$ bosons. The final linear combination remains massless - there are no more Goldstone bosons left - and forms the photon. The upshot is that after spontaneous symmetry breaking, there are now 3 massive gauge bosons, 1 massless one (that is distinct from each of the above massless gauge fields) and a Higgs boson, a "remnant" of the symmetry breaking.
To stress again, this mass generation occurs at exactly one point - the electroweak transition scale. It is not a process of a massless particle having to continuously "bombard" against the Higgs boson to gain mass or anything of that sort.
So if this "mass generation" supposedly happens at a single scale, why can processes like pair production take place at any energy? It's because these are two completely distinct phenomena. In the Higgs mechanism, we are fundamentally altering the particle content of the theory. Above the electroweak scale, there simply aren't any massive fundamental particles, and below it, there are. Pair production of mass on the other hand is simply a consequence of the mass-energy equivalence of special relativity. If there did exist massive particles above the electroweak scale, then pair production would also be possible there - they are unrelated processes.
"Mass generation happens at a single scale" should not be interpreted as "all mass is created forever, and no mechanisms like pair creation and annihilation can change this". It means that the quanta of the post-SSB fundamental fields now have a mass, whereas the quanta of the pre-SSB fields did not. These are very distinct statements.
It is seen that while only the massive fields end up interacting with the residual Higgs boson post-SSB, these interactions don't play any role in generating masses. The existence of the Higgs boson does however serve as an important verification of the Higgs mechanism.
There are of course no a priori reasons that photons and gluons should be massless gauge bosons (at ordinary energies) - these are experimentally validated through the nature of the forces that they produce. That is, the massless nature of photons makes accurate predictions of the resulting Coulomb-like electromagnetic force that we observe in nature. Indeed, there is no 100% experimental confirmation of the masslessness of the photon - we can only impose tighter and tighter upper mass bounds, as mentioned here. The massless model for gluons makes similarly accurate predictions, though the subtlety of non-abelian gauge theories means that the mediated force does not manifest as a Coulomb-like force.