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.