For a finite subset $S \subseteq \mathbb Z \setminus \{0\}$, let us say $d=\gcd S$ iff $d>0 $ , $ d|a,\forall a \in S$ and $m|a,\forall a\in S \implies m|d$.
My question is:
Does there exist a positive integer $k$ such that the following holds : for every $n>1, \exists $ a finite $X_n \subseteq \mathbb Z\setminus \{0\}$ such that $|X_n|=n$ , $k=\gcd X_n$ but $\gcd (X)\ne k$ for every proper subset $X \subsetneq X_n$ ?
It's easy to construct such a set, for any $k,n$.
Let $\{p_1,\cdots, p_n\}$ be distinct primes such that none of them divide $k$. Then define $$q_i=\frac 1{p_i}\times \prod_{j=1}^np_j$$
Thus, $q_i$ is the product of all the $p_j's$ other than $p_i$.
Now define $$x_i=q_i\times k\quad \text {and}\quad X=\{x_i\}$$
To see that this works, note first that it is clear that $k=\gcd(X)$ since $k$ is a common divisor of all the $x_i$, and none of the $p_i$ divide all the $\{x_i\}$ Moreover if $S$ is a proper subset of $X$ then we can find some $x_i\notin S$ in which case each element of $S$ is divisible by $p_ik$.