Assume there is some $n,c $ for which it is impossible for Mary. Then the residues of the $a_i$ are $1,2,\cdots,n$ in some order as there doesn t may be two residues the same. Next, if n is composite, let $n=ab$ with $b\ge a$, we could $(a+1)-1$ and $(b+2)-2$ be two factors. These are $4$ different residues, hence the total product is a multiple of $n$. If $n$ is prime, take some permutation such that $a_i \equiv -ci \pmod n$ Then the product has residue $c^n \equiv c \pmod n$ . Hence it was always possible for Mary

Can you please explain the solution in more detail, especially "If n is composite... is a multiple of n." Thanks.

If $n$ is composite, let $n=ab$ with $n-1 > b\ge a > 1$. Then $1<2<a+1 < b+2$, so they are four different residues, and Mary can take $a_1 \equiv b+2$, $a_2 \equiv 2$, $a_3\equiv a+1$, $a_4\equiv 1$, and the rest arbitrarily.