Solution

Answer: This only holds for \(n=7\).

Assume that \(n\) satisfies \(n!\mid \prod_{p<q \leqslant n}(p+q)\) and let \(2=p_{1}<p_{2}<\cdots<p_{m} \leqslant n\) be the primes in \(\{1,2, \ldots, n\}\). Each such prime divides \(n\) !. In particular, \(p_{m} \mid p_{i}+p_{j}\) for some \(p_{i}<p_{j} \leqslant n\). But

\[0<\frac{p_{i}+p_{j}}{p_{m}}<\frac{p_{m}+p_{m}}{p_{m}}=2\]

so \(p_{m}=p_{i}+p_{j}\) which implies \(m \geqslant 3, p_{i}=2\) and \(p_{m}=2+p_{j}=2+p_{m-1}\). Similarly, \(p_{m-1} \mid p_{k}+p_{l}\) for some \(p_{k}<p_{l} \leqslant n\). But

\[0<\frac{p_{l}+p_{k}}{p_{m-1}} \leqslant \frac{p_{m}+p_{m-1}}{p_{m-1}}=\frac{2 p_{m-1}+2}{p_{m-1}}<3\]

so either \(p_{m-1}=p_{l}+p_{k}\) or \(2 p_{m-1}=p_{l}+p_{k}\). As above, the former case gives \(p_{m-1}=2+p_{m-2}\). If \(2 p_{m-1}=p_{l}+p_{k}\), then \(p_{m-1}, so \(k=m\) and

\[2 p_{m-1}=p_{l}+p_{m-1}+2 \Rightarrow p_{m-1}=p_{l}+2=p_{m-2}+2\]

Either way, \(p_{m-2}>2\) and 3 divides one of \(p_{m-2}, p_{m-1}=p_{m-2}+2\) and \(p_{m}=p_{m-2}+4\). This implies \(sp_{m-2}=3\) and thus \(p_{m}=7\), giving \(7 \leqslant n<11\). Finally, a quick computation shows that \(7! \mid \prod_{p<q \leqslant 7}(p+q)\) but \(8!\nmid \prod_{p<q \leqslant 7}(p+q)\), so neither does \(9!\) and \(10!\).