4. Let $n$ be a given positive integer, and let the set $S=\{1,2$, $\cdots, n\}$. For non-empty finite sets of real numbers $A$ and $B$, find the minimum value of $|A \otimes S|+|B \otimes S|+|C \otimes S|$, where $C=A+B=\{a+b \mid a \in A, b \in B\}$, $X \otimes Y=\{x \mid x$ belongs to exactly one of $X$ and $Y\}$, and $|X|$ denotes the number of elements in the finite set $X$.
4. The sought minimum value is $n+1$.

First, take $A=B=S$, then we know
$$
|A \otimes S|+|B \otimes S|+|C \otimes S|=n+1 \text {. }
$$

Next, we prove:
$$
l=|A \otimes S|+|B \otimes S|+|C \otimes S| \geqslant n+1 .
$$

Let $X \backslash Y=\{x \mid x \in X, x \notin Y\}$. Clearly,
$$
\begin{aligned}
l= & |A \backslash S|+|B \backslash S|+|C \backslash S|+ \\
& |S \backslash A|+|S \backslash B|+|S \backslash C| .
\end{aligned}
$$

Thus, we only need to prove:
(1) $|A \backslash S|+|B \backslash S|+|S \backslash C| \geqslant 1$;
(2) $|C \backslash S|+|S \backslash A|+|S \backslash B| \geqslant n$.

First, we prove (1).
In fact, if $|A \backslash S|=|B \backslash S|=0$, then $A, B \subseteq S$.
Therefore, 1 must not be an element of $C$, i.e., $|S \backslash C| \geqslant 1$.
Next, we prove (2).
If $A \cap S=\varnothing$, then $|S \backslash A| \geqslant n$, and the conclusion holds.
If $A \cap S \neq \varnothing$, let the largest element in $A \cap S$ be $n-k(0 \leqslant k \leqslant n-1)$. Then
$|S \backslash A| \geqslant k$.
On the other hand, for $i=k+1, k+2, \cdots, n$, either $i \notin B$ (in which case $i \in S \backslash B$), or $i \in B$ (in which case $n-k+i \in C$, i.e., $n-k+i \in C \backslash S$).
Therefore, $|C \backslash S|+|S \backslash B| \geqslant n-k$.
From (1) and (2), we get (2).
In summary, $l \geqslant n+1$.
Thus, the minimum value is $n+1$.