62nd Ukrainian National Mathematical Olympiad

Let these numbers be $a_1,a_2,...,a_m$. It is enough to show that you can choose some two of these numbers $a_i,a_j$ (with $i<j$) so that all the numbers $a_1,a_2,...,a_{i-1},a_{i+1},...,a_{j-1},a_{j+1},...,a_m,a_i+a_j$ give distinct remainders when divided by $n$, then we can combine the numbers $(m-k)$ times and get the statement of the problem. If any of the numbers is divisible by $n$, for example, $a_1$ then we can combine the numbers $a_1,a_2$. Otherwise, replace these numbers with their remainders when divided by $n$ and sort them. Let $0<a_1<a_2...<a_m<n$. Then we can combine $a_1,a_m$. Indeed: for any $j$, $a_1+a_m>a_j$ and $a_1+a_m<a_j+n$, so this remainder does not occur among the others.