The answer is $\boxed{288}$. First we will prove that $288\mid P_S$ For all $S$. If four elements in $S$ have the same residue divided by $2$, then surely $32\mid P_S$. Assume that $a_2\equiv a_2\equiv a_3\not\equiv a_4\equiv a_5 \pmod{2}$. Then $2\mid a_4-a_5$, and $16\mid (a_1-a_2)(a_2-a_3)(a_3-a_1)$, so $32\mid P_S$. Also, if three elements in $S$ have the same residue divided by $3$, surely $9\mid P_S$. If not, the number of each residues mod 3 should be $2/2/1$ (not ordered). Therefore $9\mid P_S$. To show that $288$ is maximum, simply set $S=\{1,2,3,4,5\}$ then $P_S=288$.

Overkill.