Four. (15 points) The number of elements in set $S$ is denoted as $|S|$, and the number of subsets of set $S$ is denoted as $n(S)$. Given three non-empty finite sets $A$, $B$, and $C$ that satisfy the condition:
$$
\begin{array}{l}
|A|=|B|=2019, \\
n(A)+n(B)+n(C)=n(A \cup B \cup C) .
\end{array}
$$

Determine the maximum value of $|A \cap B \cap C|$, and briefly describe the reasoning process.
$$
2^{2019}+2^{2019}+2^{|C|}=2^{|A \cup B \cup C|} \text {, }
$$

that is, $2^{2020}+2^{|C|}=2^{|A \cup B \cup C|}$.
Thus, $|A \cup B \cup C|>|C|$,
$|A \cup B \cup C|>2020$.
If $|C|>2020$, then
$$
1+2^{|C|-2020}=2^{|A \cup B \cup C|-2020} \text {. }
$$

The left side of the above equation is odd, while the right side is even, so
$$
|C|>2020
$$

does not hold.
Similarly, $|C|<2020$ does not hold either.
Therefore, $|C|=2020$.
In this case, from equation (1) we know $|A \cup B \cup C|=2021$.
Clearly,
$|A \cap B \cap C| \leqslant \min \{|A|,|B|,|C|\}=2019$.
If $|A \cap B \cap C|=2019$, then $A=B \subseteq C$.
Thus, $|A \cup B \cup C|=|C|=2020$. This is a contradiction.
Therefore, $|A \cap B \cap C| \leqslant 2018$, meaning the maximum possible value of $|A \cap B \cap C|$ is 2018.
$$
\begin{array}{l}
\text { When } A=\{1,2, \cdots, 2019\}, \\
B=\{1,2, \cdots, 2018,2020\}, \\
C=\{1,2, \cdots, 2018,2019,2021\}
\end{array}
$$

then, $|A \cap B \cap C|=2018$, meaning the maximum value of $|A \cap B \cap C|$ that satisfies the given conditions is 2018.