69th Belarusian Mathematical Olympiad

Answer: $S\le 133$.

Clearly $S\le 140$. Suppose that $S\ge 134$ and let $S=134+m$, where $0\le m\le 6$. Consider the next collection of numbers: one number equals $8$, $m$ numbers equal $10$ and $14-m$ numbers equal $9$ (the total sum equals $134+m$). At least eight of these $15$ numbers will be in the same group. But the sum of the smallest eight numbers is not less than $1\cdot 8+m\cdot 9+(7-m)\cdot 10=78-m\ge 72$ — a contradiction.

It remains to show that any collection of positive integers not exceeding $10$ with the sum $S\le 133$ can be partitioned into two groups as required. We will successive put numbers to the first group (in an arbitrary order), and at some moment the sum of all numbers of this group will satisfy the next two conditions: 1) $60<A\le 70$; and 2) $A+a>70$ for any remaining number $a$. If $A\ge 63$, then $S-A\le 70$ and we can form the second group from all remaining numbers.

Now consider the case $A\le 62$. Then $A=62-x$, where $x$ equals $0$ or $1$. For any remaining number $a$ the inequality $A+a>70$ takes the form $a>8+x$, which is equivalent to $a\ge 9+x$. If there are not more than seven numbers left, their sum does not exceed $70$ and we can form the second group from them. Otherwise the sum $S-A$ of the remaining numbers is not less than $8\cdot (9+x)=72+8x$, therefore $$S=(S-A)+A\ge 72+8x+(62-x)=134+7x>133,$$ which contradicts to $S\le 133$. Thus, the required partition is constructed in all cases.