Consider the $N$ integers $1,2,\cdots,N$.
There are $\binom{N}{m}$ ways to choose $m$ distinct numbers from the set.
For example, for $N=6$ and $m=2$ we have $C_1=(1,2),C_2=(1,3),C_3=(1,4),C_4=(1,5),C_5=(1,6),C_6=(2,3),C_7=(2,4)$,$C_8=(2,5),C_9=(2,6),C_{10}=(3,4),C_{11}=(3,5),C_{12}=(3,6),C_{13}=(4,5),C_{14}=(4,6)$,$C_{15}=(5,6)$.
If I take the sum of the elements in each $C_i$, I get $S_1=3,S_2=4,S_3=5,S_4=6,S_5=7,S_6=5,S_7=6,S_8=7,S_9=8$,$S_{10}=7,S_{11}=8,S_{12}=9,S_{13}=9,S_{14}=10,S_{15}=11$.
Consider now the $\binom{\binom{N}{m}}{2}$ differences $S_j-S_i$ with $1\leq i<j\leq N$. Some of them are mutiples of $N=6$. For example, $S_{15}-S_6=S_{15}-S_3=S_{14}-S_2=S_{13}-S_{1}=S_{12}-S_1=6$ and $S_6-S_3=S_7-S_4=S_8-S_5=S_{10}-S_5=S_{10}-S_8=S_{11}-S_9=S_{13}-S_{12}=0$. If $S_j-S_i$ is divisible by $N=6$. I then set $A_{ij}=A_{ji}=1$ in an initially empty $\binom{\binom{N}{m}}{2}\times\binom{\binom{N}{m}}{2}$ matrix $A$. The above analysis shows that the symmetric matrix $A$ has $M(N=6,m=2)\equiv 12$ nonzero off-diagonal elements (the $1$'s) above the main diagonal.
A few lines of Matlab code shows that, for the case of $N=14$, $M(14,2)=252,M(14,3)=4550,M(14,4)=35301,M(14,5)=142142$,$M(14,6)=320614,M(14,7)=418950$.
These numbers are less than one tenth of $\binom{\binom{N}{m}}{2}$, showing that $A$ is indeed a sparse matrix.
My question: For general $N$ and $m$, does there exist a closed-form expression for this number $M(N,m)$?
This is a partial answer.
Let $[N] = \{1, \dots, N\}$ and for each $S \subset [N]$, let $\sum S$ be the sum of elements of $S$. For each $i = 0, \dots, N - 1$, let $$F(N, m, i) = \{S \subset [N] \mid |S| = m, \sum S \equiv i \pmod N\}$$ and let $f(N, m, i) = |F(N, m, i)|$. Now note that a pair of subsets $(S, T)$ contributes to $M(N, m)$ iff $\sum S \equiv \sum T \pmod{N}$, ie $S$ and $T$ belong to the same $F(N, m, i)$ for some $i$. Thus $$M(N, m) = \sum_i 2\binom{f(N, m, i)}{2}.$$ Also note that by definition $F(N, m, i)$ partition all $m$-sized subsets of $[N]$ so $\sum_i f(N, m, i) = \binom{N}{m}$.
The map $\{a_1, \dots, a_m\} \mapsto \{(a_1 + 1) \bmod N, \dots, (a_m + 1)\bmod N\}$ gives a bijection between $F(N, m, i) \cong F(N, m, (i + m) \bmod N)$, so $f(N, m, i) = f(N, m, j)$ if $i \equiv j \pmod m$ and therefore if $i \equiv j \pmod{\gcd(m, N)}$.
So if $N$ is coprime to $m$ then all the $f(N, m, i)$ are equal, so comparing with their sum, $f(N, m, i) = \frac1N \binom{N}{m}$. Therefore $$M(N, m) = 2N \binom{\frac{1}{N} \binom{N}{m}}{2} \sim N \left(\frac1N\binom{N}{m}\right)^2 = \frac1N \binom{N}{m}^2.$$