The answer is $n$. Let $x_1, \cdots, x_n$ be the elements of the set, and let $p_1, \dots, p_m$ be the primes that divide some elements of the sequence $\{x_i\}_{1\leq i \leq n}$ Construction: For $n=m$, let $x_i=\prod_{j\neq i} p_j$ for each $1\leq i \leq n$ This works, since for each $S\subseteq [n]$, the quantity $$\gcd_{i\in S}\{ x_i \}=\prod_{j\in [n]-S} p_j$$is unique. Bound: To prove that $m\geq n$, we proceed by contradiction and map $f:[m]\to[n]$ by asigning to each prime $1\leq j\leq m$, the index $i$ such that the element $x_i$ has minimal $\nu_{p_j}$. If $m<n$, then $f[m]$ is a proper subset of $[n]$, which contradicts the assumption: the sets $f([m])$ and $[n]$ are different, but would have common $\gcd$ equal to $$\prod_{1\leq j\leq m} p_j^{\min_{1\leq i \leq n} \nu_{p_j}(x_i) }.$$Hence necessarily $n\geq m$. $\blacksquare$