Maximum number of members in a club Some people form $n=36$ clubs. There are no two clubs with same set of club members. No two people are in the same set of clubs.
If a person is not in a club, he/she must be friend with one person in that club.
However, nobody is friend with another person in the same club (yes, a weird bunch of people).
What's the maximum possible size of a club in this setup?
This is similar to the odd town/even town problem and the Fisher inequality. I suspect we could approach in a similar linear algebra setup: https://www.cs.utexas.edu/~panni/lec20.pdf
 A: There can be at most $2^{n-1} = 2^{35}$ people in any given club $C$. That's because there are $35$ clubs other than $C$, and for every subset of those $35$ clubs (including the empty subset), at most one person can be in exactly those clubs, and also in club $C$.
As pointed out in hinkypunk's answer, we can achieve this bound. Suppose that there are $2^{36}$ people: for every subset of the $36$ clubs, there is a person exactly in those clubs. Furthermore, suppose that every person is friends with only one other person: whoever is in exactly all the clubs they are not in. Then all the conditions in the problem are satisfied, and every club contains the maximum possible amount of $2^{35}$ people.
A: My general advice is to try and solve this for small $n$ first. For me, it also helped to assume that friendship has to be an equivalence relation. Then the friendship rule becomes "every club must contain exactly one person from every equivalence class of friends", which might be easier to grasp.
Hint:
We can regard every person as a subset of the set of clubs. Since "no two people are in the same set of clubs", every person is a different subset.
So what we have to do is, find as large as possible a set of subsets of an $n$-set such that we can assign friendship in a way satisfying the rules and maximizing the largest club.
Solution:

Fortunately, we can just take all the subsets and say every set is friends with their complement set, so this setting should also give you the correct maximum size. Also note that we did not even have to assume that friendship is symmetrical nor does it matter whether a person must be friends with at least or exactly one person in every club they are not in. Of course, we also dropped the assumption that friendship has to be an equivalence relation. It just happens to be one in the solution.

