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 $i \neq j$, 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.