62nd Czech and Slovak Mathematical Olympiad

Let $z_i$ denote through the process the number of coins of the $i$-th Robin Hood.

Let $n=3$. After any step, $z_1-z_2$ modulo $3$ does not change. That means for $z_1=101$, $z_2=100$, and $z_3=99$ (out of many choices), $z_1$ will never be the same as $z_2$. Thus $n=3$ is not a solution.

Now we show that for each $n\ge 4$ any initial $z_i$ the loot can be split in equal parts.

Let $s=\sum |z_i-100|$. We decrease number $s$ as long as it can be done in the way that some of the outlaws with maximal count of coins gives his two coins to (some of the) Hoods with the minimal count of coins. If $s$ can be reduced to $0$ in this way, we are done.

If $s\ne 0$, some of the outlaws has $100-k$ coins ($k>0$), $k$ outlaws have $101$ coins each and all the others have $100$ coins each. If $k\ge 2$, we decrease $s$ in two steps: $$100-k,101,101\to 100-k+1,102,99\to 100-k+2,100,100.$$ If $k$ is even, then after $\frac{1}{2}k$ such "double" steps every outlaw will have $100$ coins each. If $k$ is odd then we end in the state in which one of the outlaws has $99$ coins, one has $101$ coins and all the others have $100$ coins. We finish as follows: $$99,100,100,101\to 99,101,101,99\to 99,102,99,100\to 100,100,100,100.$$