$1)$ 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.
Let $i$ denote each member of Blue's association and assume that there are $N$ members in total, that is, $i=1,2,\cdots, N.$ And let $j,k=1,2,\ldots, 40$ denote each of 40 commission. We will show that $N$ is at least $82$.
Consider the set
$$
S=\{(i,j,k)\;|\;1\leq i\leq N, 1\leq j<k\leq 40, i\text{ belongs to }j,k\text{-th commission.}\}.
$$ Let $d_i$ denote the number of commissions that $i$ joined. We will calculate $|S|$ using double counting method. First, note that
$$
|S|=\sum_{(i,j,k)\in S}1 = \sum_{1\leq j<k\leq 40} \sum_{i:(i,j,k)\in S}1\leq \sum_{1\leq j<k\leq 40}1=\binom{40}{2},
$$ since for each $j<k$, there is at most one $i$ in common. On the other hand,
$$
|S| = \sum_{1\leq i\leq N} \sum_{(j,k):(i,j,k)\in S}1 = \sum_{1\leq i\leq N} \binom{d_i}{2},
$$ since for each $i$, the number of pairs $(j,k)$ that $i$ joined is $\binom{d_i}{2}$.
We also have $$\sum_{1\leq i\leq N}d_i = 400,$$by the assumption.
Finally, note that the function $f(x)= \binom{x}{2} = \frac{x^2-x}{2}$ is convex. Thus by Jensen's inequality we have that
$$
\binom{40}{2}\geq |S|=\sum_{1\leq i\leq N} \binom{d_i}{2}\geq Nf\left(\frac{\sum_i d_i}{N}\right)=N\binom{\frac{400}{N}}{2}.
$$ This gives us the bound
$$
40\cdot 39 \geq 400\cdot(\frac{400}{N}-1),
$$and hence
$$
N \geq \frac{4000}{49} = 81.63\cdots
$$ This establishes $N\geq 82$. However, I'm not sure if this bound is tight. I hope this will help.
$\textbf{Note:}$ If $N=82$ is tight, then above argument implies that $d_i$'s distribution is almost concentrated at $\overline{d} = 400/82 \sim 5$.
EDIT: @antkam's answer seemingly shows that $N=82$ is in fact optimal.
Best Answer
This case can be reduced to the one with even clubs and even intersections.
Let's call the maximal number of clubs in the odd/odd and even/even cases $o_n$ and $e_n$, respectively. Gerry has already shown $e_{n-1}\le o_n\le e_{n+1}$. To see that in fact $o_n\le e_n$, consider the clubs as vectors in $\mathbb F_2^n$, indexed by the citizens, pick some club $c$, and form a set $S$ of clubs by adding $c$ to all clubs (including $c$ itself). The resulting vectors are even and have even intersections.
As shown in Bollobas' book, $e_n=2^{\lfloor n/2\rfloor}$. Thus the bounds in $e_{n-1}\le o_n\le e_n$ coincide for odd $n$, so for this case we have $o_n=2^{(n-1)/2}$. For even $n$, note that $S$ does not contain the componentwise complement of any of its elements, since the original clubs corresponding to an element of $S$ and its complement would also be complements of each other, and would thus have even intersection. Thus we can double the size of $S$ by adding the complements of all its elements, and all vectors will still be even and have even intersections (since $n$ is even). Thus in this case $o_n\le e_n/2=2^{n/2-1}$, so that $2^{n/2-1}$ is both a lower and upper bound. In summary, $o_n=2^{\lfloor(n-1)/2\rfloor}$ for either parity of $n$.