Let $\mathcal{S}$ be the set $\{1,2,3,\ldots,10\}.$ Let $n$ be the number of sets of two non-empty disjoint subsets of $\mathcal{S}.$ (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when $n$ is divided by $1000.$
1. We need to find the number of sets of two non-empty disjoint subsets of $\mathcal{S} = \{1, 2, 3, \ldots, 10\}$. Let's denote these two subsets as $A$ and $B$ such that $A \cap B = \emptyset$ and $A, B \neq \emptyset$.
2. For each element in $\mathcal{S}$, there are three choices: it can be in subset $A$, in subset $B$, or in neither subset. Therefore, there are $3^{10}$ ways to assign each element to one of these three choices.
3. However, we need to exclude the cases where either $A$ or $B$ is empty. There are $2^{10}$ ways to assign each element to either $A$ or $B$ (including the cases where $A$ or $B$ is empty). Therefore, the number of ways to assign elements to $A$ and $B$ such that neither is empty is $3^{10} - 2 \cdot 2^{10} + 1$ (we add 1 to account for the case where both $A$ and $B$ are empty, which was subtracted twice).
4. Each pair $(A, B)$ is counted twice in the above calculation (once as $(A, B)$ and once as $(B, A)$). Therefore, we need to divide by 2 to get the correct count.
5. Thus, the number of sets of two non-empty disjoint subsets of $\mathcal{S}$ is:
   \[
   \frac{3^{10} - 2 \cdot 2^{10} + 1}{2}
   \]
6. Calculate the values:
   \[
   3^{10} = 59049, \quad 2^{10} = 1024
   \]
   \[
   3^{10} - 2 \cdot 2^{10} + 1 = 59049 - 2 \cdot 1024 + 1 = 59049 - 2048 + 1 = 57002
   \]
   \[
   \frac{57002}{2} = 28501
   \]
7. Finally, we need to find the remainder when $28501$ is divided by $1000$:
   \[
   28501 \mod 1000 = 501
   \]

The final answer is $\boxed{501}$