Allow me to summarize and slightly refine the results in the current answers (if only to straighten out my own thoughts); they show that the minimum number of members $m$ satisfies $82\leq m\leq84$. They also imply strict conditions on any solution with $m=82$.
I also include my result that if $m=83$, then no member is in more than $7$ commissions. Much more can be said, but I do not have a definitive proof for the cases $m=82$ or $m=83$.
The upper bound $m\leq84$ comes from bof's construction in the projective plane of order $9$; the projective plane $\Bbb{P}^2(\Bbb{F}_9)$ consists of $91$ points on $91$ lines, with $10$ points on each line and $10$ lines through each point. Importantly, each pair of lines meets in precisely one point, and each pair of points is on precisely one line.
For $7$ distinct points in general position (no $3$ on a line, e.g. points on a smooth conic) there are precisely
$$7\times10-\binom{7}{2}\times1=49$$
lines containing these points. Removing these $7$ points and the $49$ lines containing them leaves $84$ points and $91-49=42$ lines each containing $10$ points, and any pair of lines meets in at most one point. That is, we have $84$ members in $42$ commissions, with no $2$ commissions sharing more than one member, so $m\leq84$.
The lower bound $m\geq82$ comes from Song's answer; the number of pairs of commissions that share a member is at most $\binom{40}{2}$, as there are $40$ commissions. As every commission shares at most one member, this can also be counted as the number of pairs of commissions that each member is in. If the $i$-th member is in $d_i$ commissions, then it is in $\binom{d_i}{2}$ pairs of commissions and hence
$$\sum_{i=1}^m\binom{d_i}{2}\leq\binom{40}{2}.\tag{1}$$
As there are $40$ commissions with $10$ members each, we also have $\sum_{i=1}^md_i=400$. In the inequality above we can bound the left hand side from below using the fact that for all positive integers $x$ we have
$$\binom{x-1}{2}+\binom{x+1}{2}=2\binom{x}{2}+1.$$
This allows us to even out the $d_i$'s to find that
$$\sum_{i=1}^m\binom{d_i}{2}\geq(m-n)\binom{x}{2}+n\binom{x+1}{2},\tag{2}$$
for some $x$ and $n$ with $0\leq n<m$, where
$$(m-n)x+n(x+1)=\sum_{i=1}^md_i=400.$$
The latter simplifies to $mx+n=400$, which shows that $x=\lfloor\frac{400}{m}\rfloor$ and $n=400-mx$. Plugging this back in shows that
\begin{eqnarray*}
\binom{40}{2}&\geq&\sum_{i=1}^m\binom{d_i}{2}
\geq(m-n)\binom{x}{2}+n\binom{x+1}{2}\\
&=&(m-(400-m\lfloor\tfrac{400}{m}\rfloor))\binom{\lfloor\frac{400}{m}\rfloor}{2}+(400-m\lfloor\tfrac{400}{m}\rfloor)\binom{\lfloor\frac{400}{m}\rfloor+1}{2}\\
&=&-\frac{m}{2}\lfloor\tfrac{400}{m}\rfloor^2-\frac{m}{2}\lfloor\tfrac{400}{m}\rfloor+400\lfloor\tfrac{400}{m}\rfloor.
\end{eqnarray*}
The latter is strictly decreasing for $m$ in the interval $[1,84]$. The inequality is satisfied if and only if $m\geq82$, which proves the lower bound.
Let $S$ denote the number of times we need to apply the identity $\binom{x-1}{2}+\binom{x+1}{2}=2\binom{x}{2}+1$ to reduce the left hand side of $(2)$ to the right hand side. We can then write $(2)$ more precisely as
$$\sum_{i=1}^m\binom{d_i}{2}=(m-n)\binom{x}{2}+n\binom{x+1}{2}+S.$$
Knowing that $82\leq m\leq84$ simplifies the above significantly, as then $x=\lfloor\tfrac{400}{m}\rfloor=4$ and $n=400-4m$. We find that
$$780\geq\sum_{i=1}^m\binom{d_i}{2}=1600-10m+S.$$
In particular, for $m=82$ we find that $S=0$ and hence that there are precisely $n=400-82\times4=72$ members that are in $4$ commissions and $10$ members that are in $5$ commissions. We also see that we have equality in $(1)$, meaning that every pair of commissions shares a member. This implies $\sum_{i\in C}(d_i-1)=39$ for every commission $C$, from which it follows that every commission has precisely $1$ member that is in $4$ commissions, and $9$ members that are in $5$ commissions.
If $m=83$ then $S\leq10$, and there are at most $10$ pairs of commissions that do not share a member.
Here are a few unincorporated observations that may or may not be helpful. These concern restrictions on minimal examples with $m<84$, i.e. $m=82$ or $m=83$. They are all subsumed by the observations above for $m=82$, so I prove them only for $m=83$.
Observation 1: For all $i$ we have $d_i\leq9$.
To fill the commission of member $i$ requires $9d_i+1$ distinct members, including member $i$. We have $9d_i+1\leq m=83$ and hence $d_i\leq9$.
Observation 2: For all $i$ we have $d_i\leq8$.
To fill the commission of a member $i$ with $d_i=9$ requires $9d_i+1=82$ distinct members, leaving $1$ member remaining as $m=83$. Each of the remaining $40-d_i=31$ commissions has at most one member from teach of the $d_i$ commissions of $i$, and hence contains the remaining member. But this member is in at most $9$ commissions by observation $1$, a contradiction.
Observation 3: For any pair $i$, $j$ of members in a commission we have $d_i+d_j\leq14$.
If the inequality does not hold then without loss of generality $d_i=8$ and $d_j\geq7$. To fill the shared commission requires another $8$ members, and to fill the remaining $7$ commissions of member $i$ requires another $9\times7=63$ members. Each of the $d_j-1$ remaining commissions of $j$ contains at most $7$ members from the $7$ commissions of $i$, and hence at least $2$ new members. Hence we have a total of
$$2+8+9\times(d_i-1)+2\times(d_j-1)\geq2+8+63+2\times6=85,$$
members, contradicting $m=83$.
Observation 4: For all $i$ we have $d_i\leq7$.
Suppose toward a contradiction that $d_i=8$ for some member $i$. To fill these $d_i=8$ commissions requires $9d_i+1=73$ distinct members, including member $i$, leaving $10$ members. Each of the remaining $32$ commissions has at most $8$ members from the $d_i=8$ commissions, hence at least $2$ members from the remaining $10$. Numbering these $1$ throught $10$ we find that
$$\sum_{k=1}^{10}d_k\geq2\times32=64.$$
We distinguish two cases:
Case 1: If $d_j=8$ for some $1\leq j\leq10$ then $j$ shares a commission with at least $8$ other of these $10$ members, hence they all have $d_k\leq6$ by observation $3$. To satisfy the inequality there must be one more member $j'$ with $d_{j'}=8$, and the other $8$ have $d_k=6$.
We have $11$ members, including $i$, that together take up $8+64=72$ spots in the $40$ commissions. The remaining $83-11=72$ members then take up $400-72=328$ spots. As noted before, the sum $\sum\binom{d_i}{2}$ ranging over the remaining $72$ members is minimal when the values $d_i$ differ by at most $1$. This happens precisely when $d_i=5$ for $40$ members and $d_i=4$ for $432$ members. Then
$$\sum_{k=1}^{83}\binom{d_i}{2}\geq3\binom{8}{2}+8\binom{6}{2}+40\binom{5}{2}+32\binom{4}{2}=796,$$
which exceeds the bound of $\binom{40}{2}=780$ we found before, a contradiction.
Case 2: If $d_j\neq8$ for all $10$ remaining members, then to satisfy $\sum_{k=1}^{10}d_k\geq64$ there must be a t least $4$ members with $d_k=7$. We also have $\sum_{k=1}^{10}\leq70$, and we proceed as before.
We have $5$ members, including $i$, that together take up $8+28=36$ spots in the $40$ commissions. Hence the remaining $83-5=78$ members take up $400-36=364$ spots. The sum $\sum\binom{d_i}{2}$ over the remaining $78$ members is minimized when the $d_i$ differ by at most $1$. This happens precisely if $d_i=5$ for $52$ members and $d_i=4$ for $26$ members, and we
$$\sum_{k=1}^{83}\binom{d_k}{2}\geq\binom{8}{2}+4\binom{7}{2}+52\binom{5}{2}+26\binom{4}{2}=788,$$
again contradicting the upper bound of $\binom{40}{2}=780$.
Much more can be said, but my computer is already freezing up at this big an answer.
