Provided your hands remain the same distance from the centre of each bar then in a simple physical model the weights can be any distance further out (assuming they are at the same distance and the bars remain the same weight).
Differences in arm strength (right v left) and not exactly holding the bar in symmetric positions can add lots of complicated rotations that might lead to less efficient lifting but more exercise for the muscle.
Though it is hard to be sure with physiological models, the human body is a long way from a simple mechanical model!
Analyzing the acceleration of the center of mass of the system might be the easiest way to go since we could avoid worrying about internal interactions.
Let's use Newton's second law: $\sum F=N-Mg=Ma_\text{cm}$, where $M$ is the total mass of the hourglass enclosure and sand, $N$ is what you read on the scale (normal force), and $a_\text{cm}$ is the center of mass acceleration. I have written the forces such that upward is positive
The center of mass of the enclosure+sand moves downward during process, but what matters is the acceleration. If the acceleration is upward, $N>Mg$. If it is downward, $N<Mg$. Zero acceleration means $N=Mg$. Thus, if we figure out the direction of the acceleration, we know how the scale reading compares to the gravitational force $Mg$.
The sand that is still in the top and already in the bottom, as well as the enclosure, undergoes no acceleration. Thus, the direction of $a_\text{cm}$ is the same as the direction of $a_\text{falling sand}$ . Let's just focus on a bit of sand as it begins to fall (initial) and then comes to rest at the bottom (final). $v_\text{i, falling}=v_\text{f, falling}=0$, so $a_\text{avg, falling}=0$. Thus, the (average) acceleration of the entire system is zero. The scale reads the weight of the system.
The paragraph above assumed the steady state condition that the OP sought. During this process, the center of mass apparently moves downward at constant velocity. But during the initial "flip" of the hour glass, as well as the final bit where the last grains are dropping, the acceleration must be non-zero to "start" and "stop" this center of mass movement.
Best Answer
Let's suppose you are pulling yourself up and down in approximately simple harmonic motion so your height above the ground will be give by:
$$ h = h_o + h' sin(\omega t) $$
Your acceleration is just $d^2h/dt^2$, and the force is just your mass times the acceleration, so the force due to your motion will be:
$$ F = - m h' \omega^2 sin(\omega t) $$
and the total force on the bar is:
$$ F = - m \left( g + h' \omega^2 sin(\omega t)\right) $$
So in this model the force is greatest at the bottom of your cycle as you are slowing your decent and accelerating yourself back up. It is lowest at the top where your ascent is slowing and you're allowing gravity to pull you back down.