While working on another problem, I arrived at the following, which seems interesting in its own right:
Given positive integers $n,m,k,l$ and some $a\in\mathbb{Z}/n\mathbb{Z}$, count the sets of $m$ subsets of size $k$ of $\mathbb{Z}/n\mathbb{Z}$ such that their intersection has size $l$, the sums of the $m$ subsets are all different and the sum of these sums is $a$.
For example, if $n=7$, $m=2$, $k=3$, $l=1$ and $a=2$, we would count $$\{\{0,2,4\},\{1,5,4\}\},$$ because the size of the intersection $\{0,2,4\}\cap\{1,5,4\}=\{4\}$ is $1$, the sums of the subsets are different ($6$ and $3$), and the sum of these sums is $2$.
To be honest, I have no idea how to solve this problem. Things I know so far is that constraining the union size instead of the intersection size gives and equivalent problem, in the sense that the solutions to the original problem can be easily derived from it and vice versa, that replacing $a$ by $\phi(a)$ for some automorphism $\phi$ of $\mathbb{Z}/n\mathbb{Z}$ leaves the count unchanged, and that the same goes for replacing $a$ by $a+mk$
I have managed to find a direct formula for the amount of subsets of $\mathbb{Z}/n\mathbb{Z}$ with $m$ elements which sum to a given $a\in\mathbb{Z}/n\mathbb{Z}$, but that clearly isn't strong enough.