, 2023

This only holds for $n = 7$.

Let $2 = p_1 < p_2 < \cdots < p_m \le n$ be the primes $\le n$. Each such prime divides $n!$.

In particular, $p_m \mid p_i + p_j$ for some $p_i < p_j \le n$. But
$$0 < \frac{p_i + p_j}{p_m} < 2,$$
so $p_m = p_i + p_j$ which implies $m \ge 3$, $p_i = 2$, and $p_m = 2 + p_j = 2 + p_{m-1}$.

Similarly, $p_{m-1} \mid p_k + p_l$ for some $p_k < p_l \le n$. But
$$0 < \frac{p_l + p_k}{p_{m-1}} < 3.$$
Thus, it is easy to check that we must have $p_{m-1} = p_{m-2} + 2$.

Hence, one of $p_{m-2}, p_{m-1}, p_m$ is divided by $3$. Easy to check that $p_{m-2} = 3$ is the only possible case, i.e., $p_1, p_2, \cdots, p_m = 2, 3, 5, 7$ and
$$\prod_{i<j} (p_i + p_j) = 5 \cdot 7 \cdot 9 \cdot 8 \cdot 10 \cdot 12 = 7! \cdot 60$$
So the condition only holds for $n = 7$.