This question arose in Math.StackExchange with $k=3,m=3$ https://math.stackexchange.com/questions/4179825/for-which-n-in-bbb-n-can-we-divide-1-2-3-3n-into-n-subsets-each-wi

For which $m,k$ are there infinitely many $n$ for which the numbers from $1$ to $mn$ can be split into $n$ disjoint $m$-tuples $x$ with $$kx_1=sum_{igt 1}x_i$$

Clearly $sum_{i=1}^{mn}i$ must be a multiple of $k+1$.

We also have $kle(m-1)^2$ otherwise the collection of $x_1$s can’t be small enough to match the rest of the $x_i$.

Are these two conditions on $k, m$ and $n$, along with $ngt f(m,k)$ for some $f$, enough to ensure a solution?

For example, if $k=1, m=3$ and there is a solution for $n=N$, then there is also a solution for $n=4N$: Double all the numbers in the solution for $n=N$, then include the vectors $$(9N+a,9N+1-a,2a-1),a=1..3N$$