69th Belarusian Mathematical Olympiad

Answer: $\max S=17.1$.

First we will show that if $S>17.1$, the required partition can be impossible. Indeed, let $S=17.1+19ϵ$, $ϵ>0$. Suppose that we are given $19$ numbers equal to $0.9+ϵ$. It is evident that for any partition some of the groups will contain at least $10$ of these numbers. But then the sum of the numbers in this group is not less than $9+10ϵ>9$, which is forbidden.

Now let $S\le 17.1$. We will prove that the required partition exists for any such $S$, which will imply $\max S=17.1$. Consider all sums $S_1$ of (some of) the given numbers such that $8<S_1\le 9$, and let $A$ be the maximal sum (if such sums don't exist, there is nothing to prove). For any remaining number $a$ holds $A+a>9$. Consider the next two cases:

Case 1: $A\ge 8.1$. The sum of all remaining numbers doesn't exceed $9$ and we can make the required partition into two groups with the sums $A$ and $S-A$.

Case 2: $A<8.1$. Let $A=8.1-x$, $0<x<0.1$. If there are not more than nine numbers left, then their sum doesn't exceed $9$ and we can form the second group from them. Suppose that there are at least ten numbers left. Since for any remaining number $a$ holds $A+a>9$ and $A=8.1-x$, we have $a>0.9+x$. Hence $S-A\ge 10\cdot (0.9+x)=9+10x$. But then $S\ge A+9+10x=8.1-x+9+10x=17.1+9x>17.1$ - a contradiction.