A subset here, a subset there ....

If $n = p$ a prime then if $p | ab$ then either $p | a$ or $p | b$ , (or both).

Say $p$ divides only one of the elements. Without loss of generality let this element be $a$ . There are then $x = \lfloor \dfrac{1000}{p} \rfloor$ numbers to choose from for $a$ and $1000 - x$ numbers to choose from for $b$ . This gives us $x(1000 - x)$ subsets.

If $p$ divides both $a$ and $b$ then we have $\dbinom{x}{2}$ subsets to add to the total.

Thus for $n = p$ prime, we have $S(p) = x(1000 - x) + \dbinom{x}{2}$ , where $x = \lfloor \dfrac{1000}{p} \rfloor$ .

For $p = 13$ we have $x = 76$ and $S(13) = 76*924 + \dbinom{76}{2} = 73074$ .

For $p = 31$ we have $x = 32$ and $S(31) = 32*968 + \dbinom{32}{2} = 31472$ .

Thus $S(13) - 2*S(31) = 73074 - 2*31472 = \boxed{10130}$ .