I am a student so please point out in gory detail anything I did wrong.
For a process to be quasistatic, the time scales of evolving the system should be larger than the relaxation time. Relaxation time is the time needed for the system to return to equilibrium.
We have an adiabatic process, so equilibrium must be preserved at each point, that is to say
(Working within the validity of the kinetic theory for ideal gases and ignoring friction)
$(A L(t))^\gamma P(t) = (A L_0)^\gamma P(t_0)$
Momentum gained by the piston:
$\Delta p = 2 m v_x$
A molecule would impact the piston every
$\delta t = \frac{2(L_0+ \delta x) }{v_x}$
The force exerted on the piston is $F =\frac{\Delta p}{\delta t} = \frac{m v_x^2}{L_0+\delta x}$ Pressure would be $P = \frac{P}{A}$ and for $N$ such molecules
$P = \frac{N m <v_x>^2}{A (L_0+\delta x)} = \frac{N m <v>^2}{3A (L_0+\delta x)}$
So at the instant $t=t'$ where the piston has been displaced by $\delta x$, we have
$(A L(t))^\gamma P(t) = \frac{N m <v>^2}{3A^{1-\gamma}} (L_0+\delta x)^{\gamma -1}$
Expanding in series
$ = \frac{N m <v>^2 L_0^{\gamma-1}}{3A^{1-\gamma}} (1 + \frac{(\gamma-1) \delta x}{L_0}+ O(\delta x^2) ) $
Substituing $\frac{\delta x}{L_0} = \frac{\delta t v_x}{2 L_0} -1$
$(A L_0)^\gamma P(t_0) = (A L_0)^\gamma P(t_0) (1 + (\gamma-1) (\frac{\delta t v_x}{2 L_0} -1))$
If we want our process to be reversibly adiabitic atleast to first order, we must have from above
$\delta t = \frac{2 L_0}{<v_x>}$
Now, this is time until collision for the starting case. Investigating second orders
$ (A L_0)^\gamma P(t_0) = (A L_0)^\gamma P(t_0) (1 + \frac{(\gamma-1) \delta x}{L_0}+ \frac{1}{2} (\gamma-1)(\gamma-2) (\frac{\delta x}{L_0})^2 +O(\delta x^3) ) $
Looking at just the series terms
$ 1 + (\gamma-1)\frac{\delta x}{L_0} (1 +\frac{1}{2} (\gamma -2) \frac{\delta x}{L_0}) \approx 1$
This would be true for
$\delta t = \frac{4 L_0}{<v_x>} (\frac{1}{2-\gamma} -1)$
Now, this is the "time until next collision" for a gas molecule hitting the piston. To maintain reversibility, at least to second order, the piston should be moved from $L_0$ to $L_0 + \delta x$ in time $\tau = \delta t$ so that the system variables follow the adiabatic curve.
The $<v_x>$ can be calculated from the maxwell distribution
For work to be performed, an imbalance would have to be created. That means, forces would be imbalanced and the velocity of something somewhere would have to change. A system in equilibrium won't perform work on anything.
In your case, you could add mass to your original mass and that could create an imbalance and cause the piston to compress. Or, you could remove mass and cause the piston to expand. The act of adding or removing creates an imbalance.
Also in your case, you could heat or cool the gas. That would also create an imbalance causing the piston to compress or expand.
The original system won't do anything unless an "imbalance" is imparted to it. Nothing will happen spontaneously.
The problem you are describing sounds much like the method of "Virtual Displacements" that's used frequently in structural analysis. That method considers a system, already in equilibrium, that is given a virtual displacement. Since the system is already in equilibrium, the forces within the already-stressed system perform no net work whatsoever.
Best Answer
This is a very interesting, important and at the same time subtle matter which is useful not only in thermodynamics, but other areas as well, and not everyone quite understands it.
First off, two statements that you'd better just memorize for now:
Equilibrium, Reversibility and Driving vs Opposing Forces
Let me try to give you an informal but intuitive notion of what reversibility has to do with equilbrium and balance between driving and opposing forces.
Equilibrium is achieved when a driving force in some direction is countered by an opposing force in the opposite direction. The concept of force here can be expanded to include whatever wants to steer any situation to a particular direction. Let's work with the example you gave:
The gas wants to expand (driving force), but the thousand shots exactly oppose it (opposing force). All is well and in equilibium here; nothing moves or changes.
Now imagine that you removed one shot. For a brief moment, the force in the piston due to the gas pressure would be marginally bigger than that due to the shots' weight, such that the piston would accelerate upwards a little, thus expanding the gas. This expansion would lead to a small decrease in pressure, which would eventually equal the current weight (99,9% of the original). In other words, for a brief period of time, the driving force trumps the opposing force. Then we have equilibrium again.
If your goal is, say, $p_f = \frac{1}{2}{p_i}$, you can keep doing that until only 500 lead shots are left. Each removal is one step of your process, and note that every step is a small period of disequilibrium between two equilibrium states. We can say that our example process is an approximation of a process that never leaves equilibrium (in this case, mechanical equilibrium!). Also note that our process did some work!
On the other hand, what would happen if, instead of patiently removing one lead shot at a time, we abruptly took away 500 shots? Well, suddenly the driving force (gas pressure on piston) would be twice as big as the opposing force. The piston would crazily accelerate upwards and the gas would expand rapidly. After quite some time, of course the same state would be reached: $p_f = \frac{1}{2} p_i$, $v_f = 2 v_i$. Note however, that the work done is smaller! In this case, only 500 shots were elevated to the final height, while, in the former, the same 500 shots were elevated PLUS 500 others were left along the way, at each height of each step of the process.
The former process (slow) is a better approximation of a reversible process than the latter (fast). Note that, when something is at an equilibrium, a very small push is all you need to tip the state in the direction you want, minimizing time spent in disequilibrium. You don't need to roundhouse-kick it to the desired direction. The more the driving force surpasses the opposing force, the greater the irreversibilities of the process. However, it does happen faster that way.
We can also think of an example involving thermal equilibrium. Suppose you want to raise the temperature of an object from 0 to 100 degrees. If you just put it in contact with an object at 100 degrees, you'll have a driving force disproportionately bigger than the opposing force — that is, nature is DYING to heat that cold object. However, if you used one hundred objects, each at a different temperature, to incrementing our temperature 1 degree at a time, the process would be more gentle. We stay in equilibrium more. Our newfound intuition tells us that this process is closer to a reversible one.
Conclusion
We can sum this up answering your points: