Let us pick any $n$ clubs, and call this the base set of clubs. The rest of the clubs are currently considered uncovered. We also have a (currently empty) solution set $S$, where we aim to have the final set of $n$ members who will cover all of the clubs. Repeat the following process until there are (1) no more uncovered clubs, or (2) we have repeated $n$ times.
- Pick any uncovered club.
- Apply the given condition on this set of $n+1$ clubs: the base set + the club picked in $(1)$. From the problem statement, we get a member, say $M$, that is in atleast $15$ of these clubs.
- Add $M$ to our solution set $S$, and mark as covered, all clubs not in the base set, and having $M$ as a member.
Now there are 2 cases. Either (1) we ran the above steps $k$ times, $k \leq n$ and we have no more uncovered sets. Or (2) we ran the above steps exactly $n$ times, and there are still some uncovered sets.
Case (1): The only problem here is that some of the base sets might still not be covered. Note that each of the $k$ members in $S$ are present in at least $14$ of the base sets. That accounts for at least $14k$ slots out of the available $14n$ slots in the base set. Which means atleast $14k/14 = k$ base sets have been covered, leaving us with at most $n-k$ uncovered base sets. Pick one member from each of these (at most) $n-k$ base sets and add them to our solution set $S$, and we have a valid final solution set.
Case (2): We have exactly $n$ members in our solution. Each of those $n$ members are present in atleast 14 of our base sets, thus taking up atleast $14n$ base slots, which happen to be all of our base slots. Hence if we now consider the following $n+1$ clubs: base set + any uncovered club, we get a contradiction to the given condition in the problem statement.
EDIT: It seems that the choice of $14$ and $15$ in the problem statement is arbitrary, any two consecutive numbers would have served the purpose.