69th 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 $S=11.2+14ϵ$, $ϵ>0$. Suppose that we are given 14 numbers equal $0.8+ϵ$. 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.