Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

The required value is $6(\lfloor n/2 \rfloor + 1)$, unless $n=4$ in which case it is $24$. Let $S$ be a member of $\mathcal{S}$. We first show that
$$\min_{x, y \in S, x \neq y} [x, y] \le 6(\lfloor n/2 \rfloor + 1), \quad (*)$$
unless $n = 4$. To this end, for each $x$ in $S$, choose a positive integer $m_x$ such that $n < m_x x \le 2n$ and consider the set $S' = \{m_x x : x \in S\}$.
If $|S'| < n$, then $m_x x = m_y y$ for some distinct elements $x$ and $y$ in $S$, so $[x, y] \le 2n$.
If $|S'| = n$, then $S' = \{n+1, n+2, \dots, 2n\}$. The first even number in $S'$ is $2(\lfloor n/2 \rfloor + 1)$, and the number $3(\lfloor n/2 \rfloor + 1)$ is also in $S'$ if $n=3$ or $n \ge 5$. Consequently, $(*)$ holds for $n=3$ or $n \ge 5$, and it clearly holds for $n=2$.
If $n=4$, then
$$\min \{[x, y] : x, y \in \{5, 6, 7, 8\}, x \neq y\} = 24,$$
which is the required value by the preceding.
Finally, we show that, if $1 \le i < j \le n$, then $[n+i, n+j] \ge 6(\lfloor n/2 \rfloor + 1)$. Suppose, if possible, that $[n+i, n+j] < 6(\lfloor n/2 \rfloor + 1)$. Since $[n+1, n+2] = (n+1)(n+2) \ge 6(\lfloor n/2 \rfloor + 1)$, it follows that $j \ge 3$, so $n+j \ge 2(\lfloor n/2 \rfloor + 1)$. Hence $[n+i, n+j] = 2(n+j) = m(n+i)$, where $m$ is an integer greater than $2$. If $m=3$, then $n+i$ must be an even number less than $2(\lfloor n/2 \rfloor + 1)$ which is impossible. If $m \ge 4$, then $n+i < 3(\lfloor n/2 \rfloor + 1)/2 \le n+1$ which is again impossible. This ends the proof.