# co.combinatorics – Split \${1,…,mn}\$ into \$m\$-tuples \$x\$ with \$sum_{igt 1} x_i=kx_1\$

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 $$(9N+a,9N+1-a,2a-1),a=1..3N$$