Belarus — Selection and Training Session

Answer: $5.5$.

We first show that $S \le 5.5$ (obviously, $S \le 6$). Suppose that $S > 5.5$. Then we may write $S = 5.5 + 19x$, where $0 < x \le \frac{1}{38}$. It can occur that we have $11$ numbers, $10$ of which are equal to $0.5 + 2x$ and the last one is equal to $0.5 - x$. It is clear that it is impossible to divide such numbers into two groups as required in the problem conditions.

Now prove that any $S \le 5.5$ satisfies the problem conditions. There is nothing to prove if $S \le 5$. So we may assume that $5 < S \le 5.5$. Then among the given numbers there are several numbers with the sum $a < 5$ and some number $t$ such that $a + t \ge 5$. Denote by $b$ the sum of all remaining numbers, $a + t + b = S \le 5.5$.

We have $b = S - (a + t) \le 0.5$.

If $a \ge 4.5$ then $t + b \le 1$, so we can include in the first group all numbers with sum $a$, and to the second group include all remaining numbers with sum $t + b$.

Finally, if $a < 4.5$ then $S - t = a + b < 4.5 + 0.5 = 5$, hence we can include $t$ in the first group and all other numbers in the second group.