A person will go to the town center if they can determine that they must be a member of some organization. If they can determine that there is at least one letter they did not receive, then they must be a member of some organization.
If they are not a member of any organization, then they received the membership of every organization. This allows them to do a simple proof by contradiction. They start off by assuming that they are not a member of any organization (and therefore have perfect knowledge of the situation), and then wait until something unexpected happens.
The simplest case is that a person receives no letters. Assuming that they are not a member of any organization implies there are no organizations, which they know to be false. Therefore, they must be a member of at least one organization.
Consider how this plays out with organizations consisting only of single members: A, B, and C
A gets letters about B and C, B gets letters about A and C, and C gets letters about A and B.
On day 1, as everyone received letters, nobody goes to the town center.
On day 2, A assumes that there are only two organizations — B and C. B received a letter about C, and C received a letter about B. Thus, neither of them should have gone to the town center. As neither of them did, A has no reason to believe that there are any other organizations. B and C follow similar reasonings.
On day 3, A assumes that there are only two organizations — B and C. On day 2, B would have reasoned as follows: "Suppose there is only one organization, with C as its only member. Then C should have gone to the town center on day 1. However, C did not, so I must be a member of some group. So I will go to the town center today." However, B did not go to the town center on day 2, so there must be an organization that A does not know about that A is a member of. So A will go to the town center. By symmetric reasoning, B and C will also go to the town center.
We know that it is not the case that everyone will always be able to determine that they are a member of some organization. Consider the case of organizations A, AB, ABC, and ABCD.
On day 1, A, having received no letters, goes to the town center.
Every day after that, A goes to the town center while B, C, and D all stay home. B thinks there's only one organization A, and only A going to the town center confirms that. C knows there's also an organization AB, but expects B to stay at home because B doesn't know about that one. D knows there's also an organization ABC, but knows that C and B are unaware of its existence. There will never be an event that disrupts anyone's expectations, so this is a stable situation.
There are two questions we still need to answer — will there always be at least one proper meeting, and will everyone who goes eventually be in a proper meeting?
First, let's establish a base condition - there will always be at least one person who goes to the town center. I'll use a group of A, B, C, and D to help explain my proof.
If anyone does not receive a letter, then they will go the first day. This is the only way in which someone will go on the first day.
Now assume that everyone received at least one letter (and so nobody went on the first day). Then on the second day A, assuming that she is not a part of any group and therefore knows the membership of every group, will consider every other person. If any of B, C, and D is a member of all the groups, then that person should not have received any letters and gone on the first day. But nobody went the first day, so A must not have perfect information, and therefore must be a member of some group.
If nobody goes on the second day, then A will go one level deeper in her reasoning. She first considers B — he knows about all the groups for which he is not a member. If C or D is a member of all of those groups, then B would have expected them to go on the first day. When he observed that they did not, he should have gone on the second day. Because he did not go on the second day, A knows that there must be some group that she does not know about and therefore is a member of. She follows the same reasoning for C and D.
If nobody goes on the third day, A will reach the deepest level of reasoning necessary for four people. She will again consider B's reasoning. Assuming that she knows about all the groups that exist, B knows only about groups that consist of C and/or D. So he, on the third day, would have considered C's reasoning for the second day. He would have assumed that he knew about all the groups that exist, and since he only knows about groups the have only C and D in them, C must have only known about a group of which D is the only member. So C, on the second day, would have thought that only D was a member of a group, and having not witnessed D go on the first day, should have gone on the second day. Thus, B, having not witnessed C go on the second day, should have gone on the third day. Thus A, if she does not see B go on the third day, will go on the fourth day.
In more formal terms, on day $k+1$ person $p_0\in P$ will determine if $\exists p_{i1}\in (P-\{p_0\}) \exists p_{i2}\in (P-\{p_0,p_{i1}\})...\exists p_{ik}\in (P-\{p_0,...,p_{i(k-1)}\})\forall g\in G: \{p_{i1},...,p_{ik}\}\cap g \ne \emptyset$
where $G$ is the set of all groups that $p_0$ knows about. If this condition is met, then there are no groups that do not contain at least one of the people considered. This means that on day $k+1$, $p_0$ expected each person in the chain $p_{ij}$ to expect the next person in the chain to have left on the previous day (with the last person expected to leave on the first day). Because the expectation was not met, $p_0$ will go to the town center on day $k+1$.
Everyone who goes will eventually be a part of a proper meeting:
Let $p_0$ be a person who will go to the town center. If $p_0$ is the sole member of an organization, then a proper meeting can be held immediately. Otherwise, $\exists g_0,p_1\mid p_0,p_1\in g_0$. Because $p_0$ and $p_1$ are both members of that organization, neither received a letter about it. Suppose there are no organizations of which $p_0$ is a member but $p_1$ is not. Then, when $p_0$ goes to the town center, $p_1$ will know there is at least one organization that they did not receive a letter for, and must therefore be a member of.
Otherwise $\exists g_1\mid p_0\in g_1\land p_1\notin g_1$. By our above reasoning, we know that anyone who only belongs to organizations $p_0$ does will go to the town center one day after $p_0$ at the latest.
For each person $p_i\in g_1$ who is a member of $g_i\mid p_0\notin g_i$, $p_1$ can follow the same chain of logic we used to prove that someone will eventually go. I'm having trouble writing out the formal logic for it, but here's the basic idea — $p_1$ will expect $p_i$ to expect etc. In the end it boils down to two cases - either an entire organization goes, or $p_1$ will go, and because $p_1$ was an arbitrary member of $g_0$ that means everyone in $g_0$ will go.
