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Is it true that if $A\subset \mathbb{N}\backslash\{1\}$ is a finite set with $\sum_{n\in A}\frac{1}{n}<2$ then there is a partition $A=A_1\sqcup A_2$ such that \[\sum_{n\in A_i}\frac{1}{n}<1\] for $i=1,2$?
This is not true if $A$ is a multiset, for example $2,3,3,5,5,5,5$.

This is not true in general, as shown by Sándor [Sa97], who observed that the proper divisors of $120$ form a counterexample. More generally, Sándor shows that for any $n\geq 2$ there exists a finite set $A\subseteq \mathbb{N}\backslash\{1\}$ with $\sum_{k\in A}\frac{1}{k}<n$ and no partition into $n$ parts each of which has $\sum_{k\in A_i}\frac{1}{k}<1$.

The minimal counterexample is $\{2,3,4,5,6,7,10,11,13,14,15\}$, found by Tom Stobart.

Additional thanks to: Tom Stobart