Problem: In a town with $n$ inhabitants, there are $N$ clubs. Each club has an odd number of members, and for any two distinct clubs there is an even number of common members. Prove that $n\ge N$.

We wrote a solution a few weeks ago, but using the notion of rank of a matrix we can make it cleaner.

Solution: Enumerate the inhabitants by $1,\ldots,n$, as well as the clubs by $1,\ldots,N$. Consider the $n\times N$ `club matrix' $$ A\in M_{n\times N}(\mathbb{Z}_2) $$ where $A_{ij}=1$ if person $i$ is a member of club $j$, and $0$ otherwise. Let $$ B=A^t\ \ A\in M_{N\times N}(\mathbb{Z}_2).$$ Since $\operatorname{rank}(B)\le \operatorname{rank}(A^t)\le n$, it suffices to show that $\operatorname{rank}(B)=N$. The matrix entry $$ B_{ij}=\sum_k A_{ki} A_{kj}$$ equals the number modulo 2 of common members of clubs $i$ and $j$. It is thus $1$ if $i=j$, and $0$ if $i\ne j$. In other words, $B$ is simply the identity matrix, and in particular it has rank $N$.

Q.E.D.