66th Belarusian Mathematical Olympiad

Let $S_i$ denote the arithmetic mean of all $n$ numbers written on the blackboard after $i$ moves, $M_i$ and $m_i$ be the greatest and the smallest of the written numbers. Show that if $M_i-m_i\ge 2$ then $$M_{i+1}-m_{i+1}<M_i-m_i.(1)$$ Indeed, since $M_i-m_i\ge 2$, we see that the segment $[m_i,M_i]$ along with its ends, contains at least one positive integer number. Hence, at least one of the inequalities $|m(i)-S(i)|\ge 1$ or $|M(i)-S(i)|\ge 1$ holds ($S_i\in (m_i,M_i)$). If exactly one of these inequalities holds, then $M_{i+1}-m_{i+1}=M_i-m_i-1$; if both the inequalities hold, then $M_{i+1}-m_{i+1}=M_i-m_i-2$. Hence, inequality (1) holds. Since the difference $M_i-m_i$ decreases when $i$ increases, after some moves this difference will be equal either to 0 or to 1. If $M_i-m_i=0$, then all numbers on the blackboard are equal, hence they coincide with their arithmetic mean, therefore they will not be changed ever after the $i$-th move.

If $M_i-m_i=1$, then there are only two distinct numbers on the blackboard: $m_i$ and $M_i-m_i+1$. So their arithmetic mean $S(i)$ is a number between $m_i$ and $m_i+1$, hence, by condition, the numbers on the blackboard will not be changed after the $i$-th move.