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Noted about the clarity. Will potentially amend the instructions. Congrats on solving it! If you got the accepted screen for a frame and your system was balanced, letting it continue to run for a few extra cycles will allow it to get to the point where small timing differences in the number of boxes received don't cause any bar to dip below 90%.

actually, if I let it run, some of them balance out at 80-85%. I simply delayed the excess product that causes it to come out of balance. 

how many accumulators are needed in the final solution?

I would have responded with the above immediately if my account was working.

I have now fixed my account and made a solution that involves a loop containing some accumulators (hypothetically, a box could cycle through all of them twice, but it would be extremely unlikely) and some more that loop on themselves to achieve an even number of outputs. the big loop is random enough that outputs are evenly dispersed. this is probably the easiest solution, granted that you can get the timing right to avoid collision.

Interesting solution. That is clever! It is cool to see what different players come up with to solve this.

To answer your question, it is possible to solve using just 3 accumulators.

I probably should have edited my post for this, but I should clarify that the two accumulators looping on themselves was actually to turn 6 outputs into 4