A tower with—apparently—different average weights depending on the inside activity The apparent paradox involves a tower with internally pumped water that appears to have different average weights depending on the pumping configuration. Since the system is closed, we expect the average weight to be independent of the internal activity. The question touches on how static weights and dynamic impulses from fluid storage and redirection are analyzed in classical mechanics.
Imagine a tower (a single tall vertical room) with a water reservoir and outlet at the ceiling. After falling to the floor, the water is continuously pumped back up to the reservoir from a floor pump.
Because the weight of the tower is determined only by the sum of its components, we expect a constant average weight. Here, the temporary loss of weight created by the water's freefall is exactly offset by the downward force from pumping the water up to the reservoir.
Now imagine a second experiment that seems to result in an inconsistent result. Imagine that the reservoir is no longer at the top of the tower but at the centre. And imagine that instead of letting the water fall gently, an additional pump attached to the reservoir shoots the water with equal energy in two opposite directions—toward the ceiling and the floor—so that the water that is shot towards the ceiling stops right before reaching it (without touching it) and then falls down. The following observations appear evident:

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*The speed whereby the water reaches the floor in the second experiment is the same for both flows and equal to the speed measured in the first experiment.


*Furthermore, the average freefall time should be the same, because if it is true that half of the water reaches the floor earlier, the other half is suspended for a longer time, compensating exactly for the lost time.


*Finally, as the two flows are shot in opposite directions with equal momentum, the mid-level pump does not impart any additional downward force.


*The pump at the floor, however, now needs to push the water only to the centre of the tower, and so the water column going up is halved compared to the previous experiment. Consequently, the push against the floor created by the pump is halved as well.
Thus, we have a paradox: The tower in the second experiment should have the same average weight, but it appears to weigh less because one of the downward loads has been reduced.
Why should the the tower in the second experiment not weigh less than the tower in the first experiment?
What aspect am I missing?
 A: 
Also the average freefall time will be the same, because if it is true that half of the water will reach the floor earlier, the other half will fly for a longer time, compensating exactly for the lost time. [emph. added]

The error lies in the boldface portion; the compensation is not exact.
To restate the problem: We aim to either direct water up and let it fall (simple ballistic motion) or direct it to an intermediate height and then direct it up and down in such a way that some parameters match the ballistic case. Of interest is the average weight of the complete assembly, obtained by summing the dynamic impulses and static weights.
(Let the total height be $h$ and the speed to shoot water up to reach that height $h$ be $v$, where $v=\sqrt{2gh}$ from kinematics. Let's ignore drag, as we could operate in a low-pressure environment with a nonvolatile fluid if we wished.)
The two scenarios, purported to be the same in terms of the arcing-water mass transfer but paradoxically different in terms of the average total weight, are as follows:

*

*Water is directed up to reach height $h$ and then falls the distance $h$. From kinematics, the total time for a complete cycle is $\Delta t=\frac{2v}{g}$. At steady state, a light tower containing mass $m$ of water still weighs $mg$ on average because—although some water is always in freefall, not contributing to the weight—the pump or launch zone and the impact zone each contribute large downward impulses at a cyclic rate of $\frac{\dot mv}{\delta t}$, where $\delta t$ is the time required to provide or remove the water's vertical momentum. Thus, we have an total impulse over the cycle of $2\dot m v=\dot m g\Delta t=mg$, which matches that of water sitting motionless. OK, no problem.


*Water is directed up at the same rate to reach half the original height, or $\frac{h}{2}$, so the required impulse is notably less. There, it is directed (by an additional light pump) upward and downward equally so that we obtain no additional net impulse. The speeds are adjusted to exactly match those of the first case at that intermediate height. Therefore, the downward speed $v$ at the lower impact zone must also exactly match that of the first case. It seems that the average round-trip times must be the same—relative to case 1, the downward-directed water lands somewhat earlier, whereas the upward-directed water lands somewhat later—but now the assembly seems to weigh less because the ground-level pump or launching mechanism pushes down less because its output needs to reach only $\frac{h}{2}$ instead of $h$. How can this be?
Resolution: the average round-trip times are not the same. Because the water sent downward from height $\frac{h}{2}$ in case 2 arrives at the bottom relatively quickly, its residence time provides a steady load that supplements the lower impulse of the lower pump or launching mechanism to ultimately produce a total average weight of $mg$, as expected.
To illustrate this, here's an example of the motion of two water particles for case 1 (blue, where the particles stay together) and case 2 (orange, where the particles are split by the higher pump):

Note the key elements of the thought experiment: The water is motionless at height $h$ in both cases. The water always lands at the same speed, so the total impulse matches exactly between both cases. The upward and downward impulses of the higher pump in case 2 exactly offset each other. The water is shot up at a lower speed in case 2.
Again, the resolution lies in the fact that the average landing time in case 2 (midpoint of the orange landing times) precedes that of case 1 (blue landing time). Put another way, in case 2, more water sits at the bottom of the tower waiting to be reused, and this water has a weight. The original question does not account for this aspect.
(For the numbers used here, the simple arrangement of a single pump (shown in blue) produces an downward impulse rate of $\dot{m}\sqrt{2gh}$, and the double-pump arrangement (shown in orange) produces an downward impulse rate of only $\dot{m}\sqrt{gh}$, a reduction of $(1-\sqrt{2})\dot{m}\sqrt{gh}$. But the water in the second case arrives an average of $(\sqrt{2}-1)\sqrt{\frac{h}{g}}$ earlier, for an exactly offsetting impulse rate of $(-1+\sqrt{2})\dot{m}\sqrt{gh}$.)
