Belarusian Mathematical Olympiad

Answer: $\max S = 11.2$.

First we will show that if $S > 11.2$, the required partition can be impossible. Indeed, let $S = 11.2 + 14\epsilon$, $\epsilon > 0$. Suppose that we are given 14 numbers equal $0.8 + \epsilon$. The sum of any ten of these numbers exceeds $8$ and the sum of any five of them is greater than $4$. Hence the group with sum $A$ contains at most 9 numbers and the group with sum $B$ contains at most 4 numbers. Thus, the required partition is impossible.

Now let $S \le 11.2$. We will prove that the required partition exists for any such $S$, which will imply $\max S = 11.2$. Consider all sums $S_1$ of (some of) the given numbers such that $7 < S_1 \le 8$, and let $A$ be the maximal sum (if such sums don't exist, there is nothing to prove). We will show that the sum of the remaining numbers $B = S - A$ satisfies $B \le 4$. Indeed, for $A \ge 7.2$ it is clear. Suppose $A < 7.2$, then $A = 7.2 - x$, $0 < x < 0.2$. From the maximality of $A$ it follows that for any remaining number $b$ holds $A + b > 8$, i.e. $b > 0.8 + x$. Since $A + 5 \cdot (0.8 + x) = 11.2 + 4x > S$, there are at most four numbers left. And each number doesn't exceed $1$, so $B \le 4$. Thus the required partition exists.