Three. (50 points) Let $S=\{1,2, \cdots, 2016\}$. For any non-empty finite sets of real numbers $A$ and $B$, find
$$
f=|A \Delta S|+|B \Delta S|+|C \Delta S|
$$

the minimum value, where,
$$
X \Delta Y=\{a \in X \mid a \notin Y\} \cup\{a \in Y \mid a \notin X\}
$$

is the symmetric difference of sets $X$ and $Y$, and
$$
C=\{a+b \mid a \in A, b \in B\} .
$$
If $|S \cap A|=0$, then $|S \backslash A|=2016$.
Thus, equation (2) holds.
Otherwise, assume the maximum element in $S \cap A$ is $k$, then it must be that $|S \backslash A| \geqslant 2016-k$.
Since $k \in A$, we now discuss the elements
$$
x \in\{2017-k, 2018-k, \cdots, 2016\} \text {. }
$$

If $x \notin B$, then $x \in S \backslash B$; otherwise, $x \in B$, but
$$
k+x \geqslant k+(2017-k)=2017,
$$

it must be that $k+x \in C \backslash S$.
Thus, $|S \backslash B|+|C \backslash S| \geqslant k$.
Therefore, equation (2) holds.
$$
\begin{array}{l}
\text { Hence } f_{\text {min }}=2017 \text {. } \\
\text { Three, } f_{\min }=2017 \text {. } \\
\end{array}
$$

First, take $A=B=S$, then $C=\{2,3, \cdots, 4032\}$.
Thus, $|A \Delta S|=|B \Delta S|=0$,
$$
|C \Delta S|=2017, f=2017 \text {. }
$$

Hence $f_{\min } \leqslant 2017$.
Next, we prove that $f \geqslant 2017$ always holds.
Define the difference set $X \backslash Y=\{a \in X \mid a \notin Y\}$.
Then $f=|A \backslash S|+|B \backslash S|+|S \backslash C|+|S \backslash A|+$
$$
|S \backslash B|+|C \backslash S| \text {. }
$$

We only need to prove: for any non-empty real number sets $A$ and $B$, we have
$$
\begin{array}{l}
|A \backslash S|+|B \backslash S|+|S \backslash C| \geqslant 1 ; \\
|S \backslash A|+|S \backslash B|+|C \backslash S| \geqslant 2016 .
\end{array}
$$

If $|A \backslash S|=|B \backslash S|=0$, then $A \subseteq S$, and $B \subseteq S$.
Thus, $1 \notin C$.
Therefore, $|S \backslash C| \geqslant 1$, i.e., equation (1) holds.
Otherwise, $|A \backslash S| \neq 0$ or $|B \backslash S| \neq 0$, of course, equation (1) holds.