I claim the answer is $\frac{n^2-1}{2}$. Let's assume that we pick a random distribution, where all distributions are equally probable. Thus we need to find $E[cosiness]$ Consider any distribution. Denote by $d$ the number of non-empty boxes in it. Then all stone's throw away distributions from it can be achieved by picking one of $d$ non-empty boxes and one of $n-1$ boxes, but not the same as the first one, and moving stone from the first chosen box to the second. Thus, cosiness$=d(n-1)$. Now we need to find $(n-1)E[d]$. Let $p$ be the probability that a box is empty. Then $(n-1)E[d]=(n-1)n(1-p)$. But a known problem says that if we have $m$ stones and $k$ boxes, there are exactly $C_{m+k-1}^m$ distributions. Then $p=\frac{C_{2n-1}^n}{C_{2n}^n}$. Substituting this we obtain that the answer is $\frac{n^2-1}{2}$