Answer: $\max S = 17.1$.

First we will show that if $S > 17.1$, the required partition can be impossible. Indeed, let $S = 17.1 + 19\epsilon$, $\epsilon > 0$. Suppose that we are given $19$ numbers equal to $0.9 + \epsilon$. 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 $9 + 10\epsilon > 9$, 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.