SOLVED - $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.)