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
Consider the general case that $n$ people know certain secrets such that no subset of $k$ of them know all secrets, but any subset of $k+1$ of them do know all secrets. (Here $n=8$, $k=4$, secrets are locks, and knowing a secret is having a key.)
This can be done using $\binom nk$ secrets, each labelled by a different subset of $k$ among the $n$ people, which is the set of people that does not get to know that secret (everyone else does). Then for any subset of $k$ people the corresponding secret will not be known by them, but for any set of $k+1$ people and any secret there is at least one of them who knowns the secret.
To show one cannot do with less secrets, consider any solution to this problem, and the relation between on one hand the $\binom nk$ subsets of $k$ people and on the other hand the secrets, defined by none of those people knowing the secret. This relation is "one to at least one": it is required that any set of $k$ people not know at least one secret, and if two different subsets of $k$ people do not know the same secret, then the union of those subsets, which contains at least $k+1$ people would not know that secret, and hence not all secrets, which is against the requirement. The relation is in fact a surjective map from secrets to $k$-subsets, so there are al least $\binom nk$ secrets.