Assume the flies are big enough and mean enough that they weight enough to mess with and that relative sizes are such that Browian motion and the like is not a factor. Sorta like big hummingbirds with many eyes.

With a closed top jar the scale will weigh the weight of the jar plus whatever component of the pressure due to downwash that the bottom of the far feels due to the flapping of the fly's wings plus (or whatever the correct algebraic sign is) the pressure on the underneath of the jar top due to flapping. However, since the scale will always show a weight of the jar plus the flies (closed system) I would guess that the summation of the pressures inside the top and bottom of the jar plus the measured weight-on-feet of the flies is always approximately equal to the flies' weight.

With the flies at rest and no flapping the pressures are equal and opposite. Total flly weight is weight-on-feet. As the flapping increases to takeoff the bottom pressure increases, the upper pressures decreases, the weight-on-feet of the flies decreases but the total of all of that istuff s always equal to the flies' weight.

Since this is always a closed system the size of the jar, the height of the flies above the bottom of the jar and relative sizes of the jar and flies do not effect the final answer.

However -

With an open top jar the scale will weigh the weight of the jar plus whatever component of the pressure due to downwash that the bottom of the jar feels due to the flapping of the fly's wings. In this case as long as the fly is inside of the jar the bottom of the jar will feel the downwash ( or in other words, the forces on the top of the jar are removed). The weight the scale will measure is probably proportional to anything that affects the downwash component of the flies wings on the bottom of the jar - the height the flies are above the bottom of the jar, the flies position relative to the jar mouth, and the relative sizes of the jar and flies.

Sound reasonable, if so I have some swamp land I would like to sell.............

BB's in water is easy...

You have the initial weight of jar and water = to A.
Drop in BB, as the BB is submerged the weight on the scale increases an amount equal to the same volume of water displaced by the BB. This is = to B.
After the BB is submerged and as it falls the total weight is the same as the new above weight.
After it hits the bottom the weight is increased again by the difference between the BB's weight and the same volume of water. This is = to C..

The final weight is the original weight + BB weight.
Note that B + C = BB weight.