Is it possible to have non-integral multiple of $e$ as charge? We have 2 similar balls. On one ball there is charge $e$ and on other no charge. I connect the balls using a metal wire. We know that the charge gets shared until they have the same charges. Thus we get $e/2$ charge on each ball. But this contradicts the fact that charge is quantised. Help me out to find where I am wrong. If I am right, how do you explain this?
 A: You've confused the idea of charge mixing and charge density mixing.
When you connect your two spheres, charges will flow between them until eventually they have the same charge density (e.g. start with $0$ and $\rho$ as the densities, then assuming they have the same volume, the final charge density will be $\rho/2$).
Imagine the idealised situation where you have a ball with charge $e$ because there is one unpaired electron somewhere floating about inside.  The other ball has all its electrons paired to positive charges, so no net charge.  If we bring the two balls together, there must still be one unpaired electron overall, and so the overall charge is still $e$.  However, this lonely electron now occupies more space than it did previously, and so the charge density has changed.
A: First consider the case where the "balls" are individual protons. One starts out as a hydrogen atom, with one electron, and the other starts out all by itself, with no electron. If they come close enough to each other, they can form a hydrogen molecule ion, which is like a hydrogen molecule that's missing one electron. This is analyzed in section 7.3 in Griffiths' book Introduction to Quantum Mechanics, and there's also a Wikipedia page about it. In this case, that one electron forms a single molecular orbital — a quantum wavefunction encompassing both protons, shared equally by both of them. The electron is not broken into two pieces; it is still one electron, but it is shared by the two protons in a way that requires quantum physics to describe. 
If the two "balls" are macroscopic, then in principle the electron could be in a similar quantum superposition, shared equally between the two balls in a way that only quantum physics can properly describe. However, in practice, this kind of macroscopic quantum superposition rapidly "decoheres." The result might be that the electron is localized in the wire between the two balls, or it may end up on just one of balls (so that their charges differ by $e$). The one thing that definitely cannot happen is for the electron to split into two $e/2$ charges, each localized on a different ball.  
The intuition that the two balls must end up with equal charges may be correct for most practical macroscopic purposes, but we can't expect it to always be perfect.
By the way,  the statement "charge is quantized" is correct, but technically charge is quantized in units $e/3$ rather than $e$. The up and down quarks have charges $+2/3$ and $-1/3$, respectively, times the charge of a proton. This doesn't change the answer to your question at all, though, because your question was about electrons, not about quarks; and even if it were about quarks, that would only change the unit of charge without changing the essence of the answer.
A: When it comes to electrons, electrostatic laws holding for continuous
charge distributions cannot hold without qualification. And not for
quantum reasons. In ordinary conditions you have first to take into
account thermal fluctuations.
Consider an uncharged, short-circuited capacitor. Thermal fluctuation of charge on its plates amounts to $\sigma(Q)=\sqrt{kTC}$. 
At $T=300\,\mathrm K$, taking $C=10\,\mathrm{pF}$, we find 
$\sigma(Q)=2\cdot10^{-16}\,\mathrm C$ or about 1000 electrons. 
A: The things is, charge isn't really distributed uniformly over the sphere, it must be made of atoms, which carry electrons. Usually there are so many of them that we can in practice treat the distribution as continuous, but in the end of the day, you will only be able to transfer electrons between the spheres, you can't take part of the charge of one electron and leave it with half its original charge : it will  wither be transported along with its charge, or not at all.
