QUESTION: A town with $n$ inhabitants has $m$ clubs such that each club has an odd number of members and any two different clubs have an even number of common members. Show that $m\leq n$.
ATTEMPT: We argue by contradiction. Let $M$ be the smallest value of $m$ for which the proposition is false.
So we assume that there is a town with $N$ inhabitants and $M$ clubs satisfying the conditions of the problem but still having $M>N$.
Let $G$ be a bipartite graph having $X$ and $Y$ as its partite sets. The members of $X$ are the clubs and the members of $Y$ are the inhabitants. We join a vertex $A\in X$ with a vertex $p\in Y$ iff $p$ is in the club $A$.
Now if there exists $S\subsetneq X$ such that $|N(S)|<|S| $ then consider another town which has $N(S)$ as its inhabitants and $S$ as its clubs. Then we see that this town defies the proposition but with a smaller number of clubs than $M$, contradicting the minimality of $M$. Thus we may assume that $|N(S)|\geq |S|$ for all $S\subsetneq X$.
Let $S\subseteq X$ with $|S|=M-1$. Now by Hall's Theorem, there is a matching in $G$ saturating $S$ into $Y$, giving $|Y|\geq M-1$.
So the only possibility we need to rule out is $M=N+1$ and here I am stuck.
I have not been able to make use of the facts that the degree of each vertex in $X$ is odd and that $|N(x_1)\cap N(x_2)|$ is even whenever $x_1,x_2\in X$ with $x_1\neq x_2$.
Can somebody please help me from here?
Thanks.
