Not typical for me problem

[ins-]

There are given $n$ numbers $(n \ge 2)$ that we must order in table with $k$ $(k \ge 2)$ columns in such a way that in $k-1$ columns we have equal count of numbers and in the remaining column we have at least $1$ number but the the count of numbers in it is not greater that in the remaining columns. What are the conditions for n and k for that we can do the ordering mentioned?

[Circiter]

It seems to be something from the divisibility theory (sure: all the columns, but last one, are related to integer part of some rational number, and last column -- to the number-theoretic remainder...) More precisely, I think that the desired relation may look like $n\not\equiv 0 \pmod p$, where $p=\Left[n/(k-1)\Right]$.

[ins-]

Does it work for the case n=19, k=6?

[Circiter]

Oops... My formula... it's wrong, of course. Nevertheless, without modular-arithmetic notation, we can write sufficient condition as $0<n-p(k-1)\leq p$ ($p$ is defined above.) E.g. in the case of $n=19$ and $k=6$ we have $p=3$, consequently $n-p(k-1)=19-3\cdot5=4>p$, thus, desired arrangement with such a parameters is impossible... Feel free to play with this formula, I think it's close to the true solution... :)