Yes, this is not the hardest problem ever, but here's the mechanics calculation that leads to the yes answer.
Draw a free body diagram of your body as you are standing still with one foot on each scale. You experience three forces (I will label their magnitudes): (1) The force due to gravity pulling you down, $W$ (aka your weight), (2) the normal force $N_1$ of scale 1 pushing up on one foot, and (3) the normal force $N_2$ of scale 2 pushing up on your other foot. Since your body is not accelerating, these forces balance by Newton's Second Law;
$$
W = N_1+N_2
$$
Now the question is, how are these forces related to what the scales read? Well, each scale reads the force of the corresponding leg that pushes down on it. Let's call the magnitude of these forces $W_1$ and $W_2$. As it turns out, Newton's Third Law tells us that the magnitude of the force that each scale exerts on each foot (the normal force) equals the corresponding magnitude of the force that each foot exerts on the scale;
$$
W_1 = N_1, \qquad W_2 = N_2
$$
It follows that
$$
W = W_1 + W_2
$$
as desired.
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
To shamelessly steal what James says above: the scale doesn't measure your mass, which remains the same no matter where you are, or what movements you make. The scale measures your weight, which is your mass multiplied by the acceleration due to the Earth's gravity, acting between your feet and the base of the scale.
You will measure your correct weight only if you stand on the scales without moving. As soon as you bend down, the muscles in your body that do the bending also act to pull up the lower half of your body. So this reduces the pressure your body places on the scales, and make you appear to weigh less.
Then, when you straighten up, your muscles act to force both the upper and lower halves of your body away from each other, now the scales will show a heavier weight since the lower half of your body puts a greater pressure on the scales.