USA IMO TST

The answer is $s$ composite.

Composite construction Write $s=(w+x)(y+z)$, where $w,x,y,z$ are positive integers. Let $a=wy$, $b=wz$, $c=xy$, $d=xz$. Then $$abc+abd+acd+bcd=wxyz(w+x)(y+z)$$ so this works.

Prime proof Choose suitable $a,b,c,d$. Then $$(a+b)(a+c)(a+d)=(abc+abd+acd+bcd)+a^2(a+b+c+d)\equiv 0(\mathrm{mod}s).$$ Hence $s$ divides a product of positive integers less than $s$, so $s$ is composite.

Remark. Here is another proof that $s$ is composite. Suppose that $s$ is prime. Then the polynomial $(x-a)(x-b)(x-c)(x-d)\in {\mathbb{F}}_s[x]$ is even, so the roots come in two opposite pairs in ${\mathbb{F}}_s$. Thus the sum of each pair is at least $s$, so the sum of all four is at least $2s>s$, contradiction.