Problem 1828

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.