I missed that each club contains $n$ dwarfs.

I heard there exist a solution with three or four sentences.

Is the problem right? For example, if there is only one dwarf in each street, they randomly joined the club, the conclusion isn't right.

Yes, the problem is right. Each club contains $n$ dwarfs so if there are only a few dwarfs, then in each clubs members are the same

Please edit your first post and add the conditions that each club contains $n$ dwarfs. Only the first post will be shown when viewing from the contest collections.

I edites. Any solution?

I think you missed the condition that "there are no two clubs with exactly same members".

Well, this is quite easy, at least compare to last year P3. Consider corresponding bipartite graph $G$ between set $D$ of vertices representing dwarfs and set $C$ of vertices representing clubs. By Hall's marriage theorem, we get that one of these two scenarios happens: There's matching that entirely cover $C$ to distinct vertices in $D$. By Pigeonhole, there're $n$ dwarfs that lives on the same street, done. There exists (one of smallest size) non-empty set of $W\subseteq C$ such that $|W|>|N_G(W)|$. For any vertex $w\in W$, by Hall's and definition of $W$, there exists matching that entirely cover $W-\{ w\}$ to distinct vertices in $N_G(W)$, this gives $|W|-1\leqslant |N_G(W)|\implies |W|-1=|N_G(W)|$, and also we get that, for any vertex $w'\in W-\{ w\}$, the $n$ dwarfs belong to club $w'$ is the desired dwarfs. @above I don't think that's necessary condition

@above Can you clarify the last sentence in your proof? Thanks.

ThE-dArK-lOrD wrote: For any vertex $w\in W$, by Hall's and definition of $W$, there exists matching that entirely cover $W-\{ w\}$ to distinct vertices in $N_G(W)$, this gives $|W|-1\leqslant |N_G(W)|\implies |W|-1=|N_G(W)|$, and also we get that, for any vertex $w'\in W-\{ w\}$, the $n$ dwarfs belong to club $w'$ is the desired dwarfs. For any non-empty subset $T$ of $W-\{ w\}$, we have $|T|<|W|$. So, by definition of $W$, $|N_G(T)|\geqslant |T|$, note that $N_G(T)\subseteq N_G(W)$. By Hall's, we get that there's matching that entirely cover $W-\{ w\}$ to distinct vertices in $N_G(W)$, so $|W|-1\leqslant |N_G(W)|$. When combine with $|W|>|N_G(W)|$, we get that $|W|-1= |N_G(W)|$ and hence the matching that entirely cover $W-\{ w\}$ to distinct vertices in $N_G(W)$ must be perfect matching. Consequently, for any $n$ members of any specific club $w'\in W-\{ w\}$, there are $n$ distinct corresponding clubs that they belong, while they also know each other since they are members of club $w'$.

I just realize there's even easier solution! We try to inductively select one new (i.e. not selected before) dwarf from each club. If this can be carry out for all $k(n-1)+1$ clubs, PHP gives us there're $n$ from the selected $k(n-1)+1$ dwarfs that live on the same street, and so know each other, done. On the other hand, if this process stop, i.e. certain club can't nominate new dwarfs, it means that all the $n$ dwarfs in that club have been selected. These $n$ dwarfs know each other and also come from $n$ different clubs, done.