$1)$ Every club must have an odd number of members, and
$2)$ Any two clubs must share an even number of members.
Show that it is not possible to form more than $n$ clubs.
Let $C_1, C_2, \ldots, C_m$ be $m$ clubs satisfying $1)$ and $2)$.
For each $C_i$, let $x_i$ be its $(0,1)$-characteristic vector.
The characteristic vectors are a $n$-dimensional vectors, which components corresponds to a resident, and has value $1$ if he is in the club, and $0$ if not.
The dot product $x \cdot y$ of two of these vectors is the size of the intersection of the corresponding sets.
Rules $1)$ and $2)$ can now be written
$$
\begin{align}
& 1) \: x_i \cdot x_i \:\: \text{is odd} \\
& 2) \: x_i \cdot x_j \:\: \text{is even}, \: \text{for} \: i \neq j \\
\end{align}
$$
In modulo $2$
$$
\begin{align}
& 1) \: x_i \cdot x_i = 1 \\
& 2) \: x_i \cdot x_j = 0, \:\: \text{for} \:\: i \neq j \\
\end{align}
$$
If these vectors are linearly independent there can be at most $n$, since they are $n$-dimensional.
Suppose that for some scalars $c_1, c_2, \ldots, c_m$
$$c_1 x_1 + c_2 x_2 + ... + c_m x_m = 0$$
Now, for any $i$, take the dot product of both sides with $x_i$. Because of
rule $2)$ everything cancels out, except for:
$$c_i x_i \cdot x_i = c_i = 0$$
Hence all the $c_i$'s are $0$, and the vectors are linearly independent, which proves the claim.
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.
Best Answer
Each person $p$ is a member of at most $\lfloor {n-1\over2}\rfloor$ clubs, because removing $p$ from those clubs leaves several disjoint sets of size at least $2$. Therefore, the number of pairs $(p,C)$ such that $p$ is a person, $C$ is a club, and $p$ is a member of $C$, is at most $n\cdot \lfloor \frac{n-1}2\rfloor$. On the other hand, the number of such membership pairs is at least $3k$, since each $C$ contains at least $3$ members. Therefore, we get that $$ n\cdot \left\lfloor \frac{n-1}2\right\rfloor\ge 3k\implies k\le \left\lfloor\frac{n}3 \left\lfloor\frac{n-1}2 \right\rfloor\right\rfloor $$ When when $n\equiv 1$ or $3\pmod6$, the floors disappear, and this becomes $n(n-1)/6$. For these $n$, the bound is exactly attainable using a Steiner triple system of order $n$. See [Skolem] for the details of how to construct Steiner triple systems for these $n$.
Edit 2: Okay, I found this problem in the CRC Handbook of Combinatorial Designs. It is known as a packing design. It turns out that the upper bound $\left\lfloor\frac{n}3 \left\lfloor\frac{n-1}2 \right\rfloor\right\rfloor$ is attainable, except for when $n\equiv 5\pmod 6$, in which cases you can only get that upper bound minus one. The construction for $n\equiv 0,2\pmod 6$ is easy, just take a Steiner triple system of order $n+1$ and delete one person, and all clubs containing them. For $n\equiv 4,5$, you will have to consult the handbook for the construction, in the Packings chapter (chapter 40 page 550 in the second edition).