For every $A \subset S$, let
$$
S_{\mathrm{A}}=\left\{\begin{array}{ll}
(-)^{\mid \mathrm{A}} \mid \sum_{\mathbf{a} \in \mathrm{A}} a, & A \neq \varnothing, \\
0, & A=\varnothing .
\end{array}\right.
$$

Find $\sum_{\mathrm{A} \subset \mathrm{S}} S_{\mathrm{A}}$.
Solve: There are $2^{n-1}$ sets $A$ satisfying $n \in A \subset S$, and there are also $2^{n-1}$ sets $B$ satisfying $n \notin B \subset S$. Note that $B$ covers all subsets of $\{1,2, \cdots, n-1\}$, and each $A$ satisfying $n \in A \subset S$ can be obtained by adding the element $n$ to some $B$. For each $B \subset \{1,2, \cdots, n-1\}$, let $f(B) = B \cup \{n\} = A$. Then $S_{B} + S_{f(B)} = (-1)^{|B|} \cdot n$. By Lemma 4, there are $2^{n-2}$ sets $B$ such that $2 \mid |B|$, hence there are also $2^{n-2}$ sets $f(B)$ such that $2 \mid |f(B)|$.

There are $2^{n-2}$ sets $B$ such that $2 \mid |B|$, hence there are also $2^{n-2}$ sets $f(B)$ such that $2 \mid |f(B)|$.
$$
\begin{array}{l}
\therefore \sum_{n \in A \subset S} S_{A} + \sum_{\substack{B \subset S-\{ \\ 2+|B|}} S_{B} = n \cdot 2^{n-2}, \\
\sum_{\substack{n \in A \subset S \\ 2+|A|}} S_{A} + \sum_{\substack{B \subset S \\ 2 \mid |B|-\{n\}}} S_{B} = \cdots n \cdot 2^{n-2}, \\
\end{array}
$$

Finally, $\sum_{A \subset S} S_{A} = n \cdot 2^{n-2} + (-n) \cdot 2^{n-2} = 0$.