PROVED (LEAN)
This has been solved in the affirmative and the proof verified in Lean.
- $100
If $\delta>0$ and $N$ is sufficiently large in terms of $\delta$, and $A\subseteq\{1,\ldots,N\}$ is such that $\sum_{a\in A}\frac{1}{a}>\delta \log N$ then must there exist $S\subseteq A$ such that $\sum_{n\in S}\frac{1}{n}=1$?
Solved by Bloom
[Bl21], who showed that the quantitative threshold\[\sum_{n\in A}\frac{1}{n}\gg \frac{\log\log\log N}{\log\log N}\log N\]is sufficient. This was improved by Liu and Sawhney
[LiSa24] to\[\sum_{n\in A}\frac{1}{n}\gg (\log N)^{4/5+o(1)}.\]Erdős speculated that perhaps even $\gg (\log\log N)^2$ might be sufficient. (A construction of Pomerance, as discussed in the appendix of
[Bl21], shows that this would be best possible.)
See also
[46] and
[298].
