SOLVED

Is it true that if $A\subseteq\{1,\ldots,n\}$ is a set such that $[a,b]>n$ for all $a\neq b$, where $[a,b]$ is the least common multiple, then
\[\sum_{a\in A}\frac{1}{a}\leq \frac{31}{30}.\]
Is it true that there must be $\gg n$ many $m\leq n$ which do not divide any $a\in A$?

The first bound is best possible as $A=\{2,3,5\}$ demonstrates.

Resolved by Schinzel and Szekeres [ScSz59] who proved the answer to the first question is yes and the answer to the second is no, and in fact there are examples with at most $n/(\log n)^c$ many such $m$, for some constant $c>0$.

In [Er73] Erdős further speculates that in fact \[\sum_{a\in A}\frac{1}{a}\leq 1+o(1),\] where the $o(1)$ term $\to 0$ as $n\to \infty$.

See also [784].