Finding the minimal number of members

Solution 1:

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.

Solution 2:

This is just a partial answer. I will show that $85$ members suffice; I don't know if $85$ is the minimum.

Recall that a projective plane of order $n$ exists if $n$ is a prime power: it has $n^2+n+1$ points and $n^2+n+1$ lines; each line has $n+1$ points, and there are $n+1$ lines through each point; any pair of lines meets in a unique point, and any pair of points determines a unique line.

Consider a projective plane of order $9$; it has $9^2+9+1=91$ points and $91$ lines; there are $10$ points on each line and $10$ lines through each point. A set of points is in general position if no three of the points are collinear. Note that, if we have a set of $t$ points in general position, then the lines determined by those points (taken two at a time) cover a total of at most $t+8\binom t2$ points; as long as $t\le5$ then the number of covered points is at most $5+8\binom52=85\lt91$, so we can add another point to the set and still have them in general position. Thus we can find a set $S$ of $6$ points in general position.

Let the members of the Blue's association be the $91-6=85$ points which are not in $S$. The commissions are the lines which do not meet $S$; they have $10$ members each, and any two have exactly one member in common. Finally, by the in-and-out formula, the number of commissions is $$91-\binom61\cdot10+\binom62\cdot1=46.$$

P.S. Let $m$ be the minimum possible number of members. I showed above that $m\le85$. On the other hand, I have a small improvement on your lower bound $m\ge61$.

Suppose the $i^\text{th}$ member belongs to $d_i$ commissions; then $$\sum_{i=1}^md_i=400$$ since there are $40$ commissions with $10$ members each. Moreover $d_i\le9$ since $m\le85\lt91$. Let $k=|\{i:d_i\ge5\}|$. Then $$400=\sum_{i=1}^md_i\le4(m-k)+9k=4m+5k\le340+5k,$$ whence $k\ge12$; i.e., there are at least $12$ members who are on at least $5$ commissions. Choose two members $i$ and $j$ who are on at least $5$ commissions.

Case 1. There is a commission containing both $i$ and $j$.

First, there are $10$ members on the commission which $i$ and $j$ both belong to. Next $i$ belongs to $4$ more commissions, with $36$ additional members. Finally, $j$ belongs to $4$ more commissions, each of which contains at most one member of each of the $5$ commissions containg $i$, and at least $5$ members who haven't been counted yet, for a total of $20$ new members. This shows that $m\ge10+36+20=66$.

Case 2. There is no commission containing both $i$ and $j$.

In this case a similar argument shows that $m\ge67$.

This proves that $m\ge66$. Combining this with the upper bound shown earlier, we have $$66\le m\le85.$$