**Question**
$\def\nn{\mathbb{N}}$

For any $n \in \nn^+$, is there a finite set $S \subset \nn^+$ such that $\sum_{k \in S} \frac{1}{k} = n$ but $\sum_{k \in T} \frac{1}{k} \notin \nn^+$ for any $T \subsetneq S$?

**Background**

This question was inspired by *pallab1234*'s question on Math SE where he asked a similar question for $n = 2$ but with $S$ being a finite sequence (not necessarily distinct) rather than a set. I asked whether there is a solution for distinct reciprocals, and later found an ad-hoc solution. *pallab1234* then asked a follow up question at Subsequence and integers as a sum of $\frac{1}{n}$ but still not requiring distinct reciprocals. Like *Greg Martin* said in a comment, I believe the answer is yes (and that there are infinitely many solutions for every $n$) but do not see how this can be shown. Presumably there might be a way to construct a systematic procedure to select the leftover reciprocals of non-primes so that the later ad-hoc modulo arguments can be generalized.