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.