Even if some pattern was found in those experiments it still wouldn't change the fact that electroneutrality itself still holds since by definition it is stated for liquids **as a whole** (I keep bolding it to emphasize this point!). You *can't* violate it by subdividing a liquid, because the principle says nothing about what will happen when you do that.

OK, now that it is obvious we agree when it comes to the general idea I will explain why you are wrong

Imagine we have started with the solution - volume V - of water.

This solution contains V*10

^{-7} moles of H

^{+} and exactly the same amount of OH

^{-} (that's not true, but it will became obvious why later).

Now, we divide the solution pouring exactly half of it to the other vessel.

Most of the solution was divided equally, and most of the volume was already neutral, but every ion that was close to the border through which the solution was divided, was either left in the original vessel, or has been poured to the new one with probablity 0.5. p=0.5 - probability that H

^{+} is in the first vessel, q=0.5 - probability that H

^{+} is in the second vessel. p+q=1. And you know what? This is Bernoulli distribution. That means that - while statstically ions have been divided exactly in half - in reality they were not, as the variance in this case is something like pq. Thus both solutions are charged. Charges we are talking about are pretty small, as number of ions involved was pretty small when compared with the total number of ions in the solution, but nonetheless they exist and their presence is an inevitable result of the fact that world is in fact granular, not continuous.

Now it should be obvious why the original solution has been already charged - it was prepared by splitting other solution.