Problem 739

Solution. We will prove that the smallest such number $c=\frac{1}{n+1}$. First, if $a_{1}=a_{2}=\ldots=a_{n}=\frac{1}{n+1}$, then the possible sums are $\frac{1}{n+1}, \frac{2}{n+1}, \ldots, \frac{n}{n+1}$, which means that for any $c$ smaller than $\frac{1}{n+1}$, the statement does not hold. Now we will show that for $c=\frac{1}{n+1}$, the statement does hold. Let $a_{1}, a_{2}, \ldots, a_{n}$ be arbitrary real numbers. Consider the sums $s_{i}=a_{1}+a_{2}+\ldots+a_{i}$ $(1 \leq i \leq n)$. If any of these sums has a fractional part of at most $\frac{1}{n+1}$ or at least $\frac{n}{n+1}$, then we are done, since the distance of such a sum from the nearest integer is at most $\frac{1}{n+1}$. If, however, none of them do, then by the pigeonhole principle, at least one of the intervals

$$\left[\frac{1}{n+1}, \frac{2}{n+1}\right],\left[\frac{2}{n+1}, \frac{3}{n+1}\right], \ldots,\left[\frac{n-1}{n+1}, \frac{n}{n+1}\right]$$

contains two sums, say $s_{i}$ and $s_{j}$ (where $i<j$). Then the distance of the sum $s_{j}-s_{i}=a_{i+1}+\ldots+a_{j}$ from the nearest integer is at most $\frac{1}{n+1}$. This completes the proof of the statement.