Problem 995

[Proof] For each set $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ that meets the requirements of the problem, let $S_{A}=$ $\sum_{i=1}^{n} \frac{1}{a_{i}}$.

For any two natural numbers $a$ and $b$, at most 3 numbers $2 a-b, \frac{a+b}{2}, 2 b-a$ can form an arithmetic sequence with $a, b$. Let
$$M=\left\{1,2, \cdots, n+3 C_{n-1}^{2}\right\} .$$

It is easy to see that for any two elements in $M$, a third element can be found that does not form an arithmetic sequence with these two elements; for these 3 elements, a fourth element can be found that does not form an arithmetic sequence with any two of the 3 elements; $\cdots$; finally, when there are $n-1$ elements, a $n$-th element can be found that does not form an arithmetic sequence with any two of the $n-1$ elements. This shows that there exists a subset of $n$ natural numbers in $M$ that meets the requirements of the problem. There are only a finite number of such $n$-element subsets, so there must be one that maximizes the sum of the reciprocals of its elements, denoted as $A \subset M$.

For any $n$-element set $B$ that meets the requirements of the problem, if $B \subset M$, then $S_{B} \leqslant S_{A}$. If $B-M \neq \varnothing$, then $|B \cap M|<n$. Since $|M|=n+3 C_{n-1}^{2}$, the elements in $B-M$ can be sequentially replaced by elements in $M$. Clearly, each replacement increases $S_{B}$. Thus, $S_{B} \leqslant S_{A}$. This shows that for all $n$-element sets that meet the requirements, $S_{A}$ is the maximum.