For any finite set $X$, let $| X |$ denote the number of elements in $X$. Define
\[S_n = \sum | A \cap B | ,\]
where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\left\{ 1 , 2 , 3,  \cdots , n \right\}$ with $|A| = |B|$.
For example, $S_2 = 4$ because the sum is taken over the pairs of subsets
\[(A, B) \in \left\{ (\emptyset, \emptyset) , ( \{1\} , \{1\} ), ( \{1\} , \{2\} ) , ( \{2\} , \{1\} ) , ( \{2\} , \{2\} ) , ( \{1 , 2\} , \{1 , 2\} ) \right\} ,\]
giving $S_2 = 0 + 1 + 0 + 0 + 1 + 2 = 4$.
Let $\frac{S_{2022}}{S_{2021}} = \frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p + q$ is divided by
1000.


Let's think step by step and output the final answer within \boxed{}.
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