TITLE: Finding the minimal number of members
QUESTION [16 upvotes]: I've been working on the following problem

For every issue in the Blue's association, a commission with 10 members (belonging the Blue's) is formed in order to solve the problem. The only condition is

There can't be two commissions having more than one member in common

The Blue's association has formed this year 40 commissions.
What's the minimal amount of members in the Blue's association?

I've only found out the following

For any commission you can form $\binom{10}{2}=45$ different pairs and none of them can appear in another commission.
Since 40 different commissions are formed, the minimal number of pairs is $45\times 40=1800$.
Denote by $n$ the number of members. Thus $$\binom{n}{2}≥1800\Rightarrow n>60$$

$$$$

The minimal amount of members has to be 100 or less.

You can observe a distribution for 100 members here

My question:
Is 100 the answer or is there an ever smaller possible amount of members?
If so, how can I prove it?

REPLY [5 votes]: This post exhibits a solution with $82$ members.  Combined with the excellent answer by @Song, this means $82$ is indeed optimal.
Motivation: The excellent answer by @Song and the followup comments by @Servaes make me wonder... perhaps if we look for 41 commissions (not 40) then there is a solution with a great deal of symmetry:

(a) 82 members (the optimal answer we seek)
(b) 41 commissions (exceeds OP requirement)
(c) each member associated with exactly 5 commissions (not part of OP)
(d) each commission associated with exactly 10 members (equals OP requirement)
(e) each 2 members have exactly 1 common commission (not part of OP)
(f) each 2 commissions have exactly 1 common member (exceeds OP requirement)

This would be like a finite projective plane, but with 82 points and 41 lines.  However, in a finite projective (respectively: affine) plane, the no. of points and no. of lines are equal (respectively: almost equal), and this is probably why a solution based on FPP only reaches 84.  So I decided to look at related structures called Block Designs, Steiner Systems, etc. where there are typically many more "lines" than "points".  After quite a bit of digging, I think I found the right structure:
The solution:  It is a Steiner $S(t=2,k=5,n=41)$ system.  A Steiner system is defined by the following properties:

there are $n=41$ objects (these are the commissions)
there are $b$ blocks (these are the members), each block (member) being a subset of objects (i.e. the commissions he/she is associated with)
each block has $k=5$ objects (each member is associated with 5 commissions)
every $t=2$ objects is contained in exactly 1 block (every 2 commissions have exactly 1 common member)

So this already satisfies (b), (c) and (f).  Next, quoting from https://en.wikipedia.org/wiki/Steiner_system#Properties we have:

$b = {n \choose t} / {k \choose t} = (41 \times 40) / (5 \times 4) = 41 \times 2 = 82$, satisfying (a)
$r = {n-1 \choose t-1} / {k-1 \choose t-1} = 40 / 4 = 10$, where $r$ denotes "the number of blocks containing any given object", i.e. the number of members associated with any given commission, satisfying (d).

Thinking more, I don't think (e) can be satisfied.  However, (e) is not needed for the OP, so it doesn't matter.
So finally we just need to prove that such a Steiner $S(t=2,k=5,n=41)$ system exists.  This existence is non-trivial, but luckily more digging reveals: 

https://math.ccrwest.org/cover/steiner.html has a list of Steiner systems that are known to exist.  $S(2,5,41)$ (the webpage sometimes lists the 3 parameters in different order) is not part of any infinite families listed, but if you go further down the page it is listed as a standalone example; clicking on that link goes to...
https://math.ccrwest.org/cover/show_cover.php?v=41&k=5&t=2 which exhibits the system, created via "Cyclic construction" whatever that means.

I did not check the numbers thoroughly, but if I understand the webpage correctly, there should be 82 rows (members / blocks), each containing 5 numbers (commissions), all the numbers being 1 through 41 inclusive (the 41 commissions), each number (commission) should appear in 10 rows, and every pair-of-numbers should appear in 1 row.
I am not an expert in any of these, so if I had a mistake or misunderstanding above, my apologies.  Perhaps someone more expert can check my work?