The site's second question fails for every $n\ge2$. For $A=\{m:n/2<m\le n\}$ the pairwise least common multiples exceed $n$ (a common multiple of two distinct elements is at least twice the larger one), and every $m\le n$ divides some element, since for $m\le n/2$ some multiple of $m$ lies in $(n/2,n]$; so no $m\le n$ divides no element, and the answer no holds for a trivial reason (an observation of this page, confirmed by computation for $2\le n\le200$). The failure covers every $n$, so it is not a boundary failure; it is a misprint that the poser's own words contradict. The site's phrase copies Erdős's sentence in [Er73] (printed p. 135: "I thought that (14.1) implies the existence of an absolute constant $c$ so that there are $cn$ integers $m\le n$ which do not divide any of the $a$'s. To my great surprise this was disproved by Schinzel and Szekeres."). His other statement of the same question, [Er80] p. 111, differs only in the failing element: it counts, as $A'(x)$, the integers up to $x$ divisible by none of the $a$'s, and records his 1940 conjecture $A'(x)>cx$ for sequences $1\le a_1<\cdots<a_k\le x$ with pairwise least common multiples above $x$. Both texts report that Schinzel and Szekeres disproved the question, and their construction ([ScSz59] p. 228) counts the integers divisible by no element, so the report is true of that form and not of the printed one, which no construction is needed to refute. The change replaces "which do not divide any $a\in A$" by "which are not divisible by any $a\in A$", the counting of [Er80], and inserts "if $1\notin A$,": in the non-multiples form the set $A=\{1\}$ meets the hypothesis vacuously and leaves no such integer, because $1$ divides every $m$. The element $1$ is the one value at which no set can meet the conclusion: a set with $1$ in it is $\{1\}$, while a one-element set $\{a\}$ with $a\ge2$ leaves $n-\lfloor n/a\rfloor\ge n/2$ non-multiples. That exclusion is this page's own correction; the formal-conjectures statement of the second question (erdos_542.parts.ii, under Formalization) adds the same hypothesis $1\notin A$ for the same reason and counts with the site, and the sets of [ScSz59] exclude $1$. The defect is already in the poser's text: [Er73] prints the divisor form, and the site copies it. The form rests on these sources alone, not on which results settle it. The only result about the site's wording is the observation above; it settles no instance of the corrected Statement and counts for nothing. The first question is unchanged. The problem's standing judges the corrected Statement.