Answer: 501

Solution:

Count ordered pairs $(A, B)$ of disjoint subsets first: each of the $10$ elements goes in $A,$ in $B,$ or in neither, for $3^{10}$ pairs. Among these, $2^{10}$ have $A$ empty and $2^{10}$ have $B$ empty, with the pair $(\varnothing, \varnothing)$ counted in both, so $$3^{10} - 2 \cdot 2^{10} + 1 = 57002$$ ordered pairs have both subsets non-empty.

Disjoint non-empty subsets are never equal, so each set $\{A, B\}$ is counted twice, giving $n = \frac{57002}{2} = 28501.$ The remainder mod $1000$ is $501.$