69th Belarusian Mathematical Olympiad

Answer: $S\le 152$. Clearly $S\le 160$. Suppose that $S\ge 153$ and let $S=153+m$, where $0\le m\le 7$. Consider the next collection: $m$ numbers equal $10$ and $17-m$ numbers equal $9$ (the total sum equals $153+m$). At least nine of these $17$ numbers will be in the same group. But the sum of nine smallest numbers is at least $81$ — a contradiction.

It remains to prove that any collection of positive integers not exceeding $10$ with the sum $S\le 152$ can be partitioned as required. We will successive put numbers to the first group (in an arbitrary order), and at some moment the sum $A$ of all numbers of this group will satisfy two conditions: 1) $70<A\le 80$; and 2) $A+a>80$ for any remaining number $a$. If $A\ge 72$, then $S-A\le 80$ and we can form the second group from all remaining numbers. If $A<72$, then $A=71$ and the inequality $A+a>80$ implies that all remaining numbers equal $10$. There are not more than eight numbers left (otherwise $S\ge 71+9\cdot 10=161$, since $S\le 152$). Hence we can form the second group from them.