In contrast with the previous incorrect answers that I hadn't noticed, there isn't any ambiguity or confusion about the Bose-Einstein or Fermi-Dirac statistics for composite systems such as atoms.
A particle – elementary or composite particles – that contains an even number of elementary (or other) fermions is a boson; if it contains an odd number, it is a fermion. Bosons always have an integer spin; fermions always have a half-integer spin. By the spin-statistics theorem, the wave function of two bosons is invariant under their exchange but it is antisymmetric under the exchange of two identical fermions. These two rules are theorems for elementary particles and assuming this theorem, it's also trivial to prove these statements for composite particles.
In particular, for a neutral atom, the numbers of protons and electrons are equal so their total parity is even. That's why only neutrons matter. An isotope with an even number of neutrons is a boson (the whole atom: e.g. helium-4); an isotope with an odd number of neutrons is a fermion (the whole atom: e.g. helium-3).
That doesn't mean that composite bosons exhibit all the same physical phenomena like superfluidity that we may observe with some bosons.
The magnetic moment of a charged "elementary enough" particle scales like $1/m$ where $m$ is the mass of the particle. That's why the magnetic moment of the protons, neutrons, and nuclei are about 1,000-2,000 times smaller than the magnetic moments of the electrons. That's why the nuclear spins are largely negligible for the behavior of the atom in a magnetic field.
This is no contradiction because the whole atoms have a much larger magnetic moments than the nuclei separately – because of the neutrons: atoms are not "elementary enough" in this definition. Both the electrons' spins and their orbital angular momentum contribute to an atom's magnetic moment. Also, there exist a higher number of states because an atom is a typical example of the "addition of several angular momenta". The tensor product Hilbert space may be decomposed as a direct sum of Hilbert spaces with fixed values of the total angular momentum. The degeneracy and the magnetic moment of these components depend on the total angular momentum i.e. on the relative orientation of the nucleus-based and electron-related angular momenta.
In effect, the spectral lines of the whole atom exhibit the so-called hyperfine structure. Up to some approximation, the nuclear spin may be totally ignored. But when the considerations from the previous paragraph are properly account for, each spectral line is actually split to several nearby (3 or so orders of magnitude closer to each other) finer spectral lines, each of which corresponds to a different value of the total angular momentum of the whole atom (or, equivalently, a different value of $\vec J_{\rm electrons}\cdot \vec J_{\rm nucleus}$).
In an atom, the electron is not just spread out evenly and motionless around the nucleus. The electron is still moving, however, it is moving in a special way such that the wave that it forms around the nucleus keeps the shape of the orbital. In some sense, the orbital is constantly rotating.
To understand precisely what is happening lets calculate some observables. Consider the Hydrogen $1s$ state which is described by
\begin{equation}
\psi _{ 1,0,0} = R _1 (r) Y _0 ^0 = R _{1,0} (r) \frac{1}{ \sqrt{ 4\pi } }
\end{equation}
where $ R _{1,0} \equiv 2 a _0 ^{ - 3/2} e ^{ - r / a _0 } $ is some function of only distance from the origin and is irrelevant for this discussion and the wavefunction is denoted by the quantum numbers, $n$, $ \ell $, and $ m $, $ \psi _{ n , \ell , m } $. The expectation value of momentum in the angular directions are both zero,
\begin{equation}
\int \,d^3r \psi _{ 1,0,0 } ^\ast p _\phi \psi _{ 1,0,0 } = \int \,d^3r \psi _{ 1,0,0 } ^\ast p _\theta \psi _{ 1,0,0 } = 0
\end{equation}
where $ p _\phi \equiv - i \frac{1}{ r } \frac{ \partial }{ \partial \phi } $ and $ p _\theta \equiv \frac{1}{ r \sin \theta } \frac{ \partial }{ \partial \theta } $.
However this is not the case for the $ 2P _z $ state ($ \ell = 1, m = 1 $) for example. Here we have,
\begin{align}
\left\langle p _\phi \right\rangle & = - i \int \,d^3r \frac{1}{ r}\psi _{ 1,1,1} ^\ast \frac{ \partial }{ \partial \phi }\psi _{ 1,1,1} \\
& = - i \int d r r R _{2,1} (r) ^\ast R _{ 2,1} (r) \int d \phi ( - i ) \sqrt{ \frac{ 3 }{ 8\pi }} \int d \theta \sin ^3 \theta \\
& = - \left( \int d r R _{2,1} (r) ^\ast R _{2,1} (r) \right) \sqrt{ \frac{ 3 }{ 8\pi }} 2\pi \frac{ 4 }{ 3} \\
& \neq 0
\end{align}
where $ R _{2 1} (r) \equiv \frac{1}{ \sqrt{3} } ( 2 a _0 ) ^{ - 3/2} \frac{ r }{ a _0 } e ^{ - r / 2 a _0 } $ (again the particular form is irrelevant for our discussion, the important point being that its integral is not zero). Thus there is momentum moving in the $ \hat{\phi} $ direction. The electron is certainly spread out in a "dumbell" shape, but the "dumbell" isn't staying still. Its constantly rotating around in space instead.
Note that this is distinct from the spin of an electron which does not involve any movement in real space, but is instead an intrinsic property of a particle.
Best Answer
Angular momentum is a vector, not a scalar. In the case of hydrogen, this makes it possible for the total spin to be either 1/2+1/2 or 1/2-1/2.
By the way, in most cases we don't care at all about the total spin of an atom, because the hyperfine coupling is very weak. We care about the total spin of the nucleus, and the total spin of the electrons.
It is in general beyond the state of the art to predict with perfect reliability, from first principles, the ground-state spins of all nuclei. I'm a nuclear physicist, not an atomic physicist, but I believe the prediction of the total electronic spins is much more tractable, and can basically be done using Hund's rules.
In nuclear physics, for an even-even nucleus, the total spin is always zero. For an odd nucleus, if you know the shell that the odd nucleon is in, then the rule is that all the angular momenta couple so as to cancel, except for that of the odd nucleon. I.e., angular momentum is minimized, which is the opposite of Hund's rules. In an odd-odd nucleus, you have to deal with the coupling between the two odd spins.