Actually, every collection of $n$-element sets has a set with the largest sum of the reciprocals in the collection.

Probably the point of the question is that "largest" should mean "strictly larger than all others."

JBL, can you really show that much in the original problem???

Sorry, fedja, I don't understand your comment.

I agree with JBL, else the question is boring. Let $H(A)=\textstyle\sum_{a\in A} a^{-1}$. We can assume that if $a_1<a_2<...<a_n$ then $a_i\le 2^{i-1}$ (easy by induction). There are only a finite number of sets $A$, therefore one must attain the maximum.

The comment was that it is easy to show that the maximum is attained, but how on Earth do you show that it is unique?

Ah, all is clear . I don't have any idea how to solve the problem as I interpreted it, but it's the only other interpretation I can see for the problem, and since your interpretation makes the problem trivial and over-prescribed, I thought I'd mention the other option . (Well, it's not true that I don't have any idea whatsoever, but the idea I do have probably doesn't work: we only need to consider the lexicographically minimal $n$-sets with no three-term arithmetic progression. One might hope that these are easy to describe (and for $n \leq 5$ at least, they are, since there is a unique minimal element).)

JBL wrote: ...we only need to consider the lexicographically minimal $n$-sets with no three-term arithmetic progression. One might hope that these are easy to describe If I understand correctly, the first $n$ elements of OEIS A003278 is precisely the set you're referring to. Whether or not this set gives the greatest sum of reciprocals is unclear, though. If it were so, it would imply a stronger version of the Erdős-Turán conjecture for $k=3$ (since the reciprocal sum of that set is finite).

Sorry, I didn't mean lexicographic order, I meant the coordinatewise order on $\mathbb{Z}^n$ inherited by this set of $n$-tuples. (I don't mean to conjecture that the greedy algorithm settles Erdős-Turán .) In particular, in this partial order it's clear that if $\bf a$ is below $\bf b$ then the sum of the reciprocals of the elements of $\bf a$ is larger than the sum of the reciprocals of the elements of $\bf b$ (so we don't have to prove Erdős-Turán to prove the problem this way) but now there may be several minimal elements to check (but only finitely many).