58th Ukrainian National Mathematical Olympiad

Firstly we provide an example of such $N$ numbers. Consider numbers $a_k=p_k{p}_{k+1}{p}_{k+2}\dots p_{k+N-1}$, where $p_1,p_2,\dots ,p_N$ are $N$ distinct prime numbers and $p_{N+i}=p_i$ ($1\le i\le N-1$). Indeed, GCD of any set will include $p_k$ exactly in a power 1 if and only if $a_k$ belongs to this set. It means that GCD of all sets will be distinct.

Let us prove that it could not be less than $N$ prime divisors. Assume that numbers $a_1,a_2,...,a_N$ satisfy the condition and their product has only $m<N$ distinct prime divisors $p_1,p_2,...,p_m$. For every $p_i$ we choose $a_{k_i}$ that includes $p_i$ in the smallest possible power. Consider the set of all such $a_{k_i}$ taken once (it might be that $a_{k_i}=a_{k_j}$ with $i\ne j$, in this case still our set will include $a_{k_i}$ exactly one time). This set will consist of not more than $m<N$ numbers. GCD of all numbers of this set will include $p_i$ in the power that is equal to the smallest power in which $p_i$ belongs to $a_1,a_2,...,a_N$. Thus if we add any of the $a_i$ to our set, GCD of all numbers in the set will be the same. And we have a contradiction.