Oddtown

Idea. The dot product mod $2$ of incidence vectors counts the size of the intersection mod $2$.

Setup. For clubs $A_1,\dots,A_m$ let $v_i\in\mathbb{F}_2^n$ be the incidence vector. Then
$$v_i\cdot v_j=|A_i\cap A_j|\bmod2=\begin{cases}1&i=j,\\0&i\ne j.\end{cases}$$

Independence. Suppose $\sum_ic_iv_i=0$ with $c_i\in\mathbb{F}_2$. Taking the dot product with $v_j$ gives $c_j=0$ for all $j$. The $m$ vectors are linearly independent in an $n$-dimensional space, hence $m\le n$.

Construction. Take the $n$ clubs $\{1\},\{2\},\dots,\{n\}$.

Remark. In "Eventown" (even sizes, even intersections) the answer is $2^{\lfloor n/2\rfloor}$.

Lesson. Translate set conditions into inner products over $\mathbb{F}_2$ and use dimension counting.