Solution:

Induction on $n$.

For $n = 1$, $\{1\}$ is obviously maximal.

Now suppose $a_1 < a_2 < \ldots < a_n$ is a maximal set for $n$.

Take $a_{n+1}$ to be the smallest integer $> a_n$ such that $\{a_1, a_2, \ldots, a_{n+1}\}$ has no three members in arithmetic progression.

Now consider the sequences $b_1 < b_2 < \ldots < b_n$ which have no three in arithmetic progression and $b_{n+1} \leq a_{n+1}$. There are only finitely many such sequences. So we can find one which is maximal. Suppose it is $c_1 < c_2 < \ldots < c_{n+1}$.

Now take whichever of $a_i$, $c_i$ has the larger sum of inverses. It is clearly maximal with respect to sequences whose largest member is $\leq a_{n+1}$.

Suppose we have a sequence $x_1 < x_2 < \ldots < x_{n+1}$ with no three in arithmetic progression and $x_{n+1} > a_{n+1}$. Then we have $1 / x_{n+1} < 1 / a_{n+1}$ and, by induction, $1 / x_1 + \ldots + 1 / x_n \leq 1 / a_1 + \ldots + 1 / a_n$, so $1 / x_1 + \ldots + 1 / x_{n+1} < 1 / a_1 + \ldots + 1 / a_{n+1}$, so it is worse than the sequence we have chosen.