Four. (50 points) Let $a_{1}, a_{2}, \cdots, a_{20}$ be 20 distinct positive integers, and the set $\left\{a_{i}+a_{j} \| \leqslant i, j \leqslant 20\right\}$ contains 201 distinct elements. Find the minimum possible number of distinct elements in the set
$$
\left.|| a_{i}-a_{j}|| 1 \leqslant i, j \leqslant 20\right\}.
$$
The minimum number of elements in the given set is
100.
Example: Let
$$
\begin{array}{l}
a_{i}=10^{11}+10^{i}, \\
a_{10+i}=10^{11}-10^{i}(i=1,2, \cdots, 10) .
\end{array}
$$

Then $\left\{a_{i}+a_{j} \mid 1 \leqslant i \leqslant j \leqslant 20\right\}$ contains
$$
(20+19+\cdots+1)-10+1=201
$$

different elements.
$$
\begin{array}{l} 
\text { And }\left\{\left|a_{i}-a_{j}\right| 11 \leqslant i, j \leqslant 20\right\} \\
=\left\{2 \times 10^{i} \mid i=1,2, \cdots, 10\right\} \cup \\
\left\{110^{i} \pm 10^{j} 111 \leqslant i, j \leqslant 20\right\}
$$

Consider the pairs $(b, c)$ (a total of 190 pairs) in $S$ such that $b > c$. By the assumption, there are at most 99 different differences, so there must be at least 91 pairs $(b, c)$ such that there exist $b^{\prime}, c^{\prime} \in S$ satisfying $b^{\prime}<b, c^{\prime}<c$, and $b-c=b^{\prime}-c^{\prime}$. For these 91 pairs $(b, c)$, they and their corresponding $b^{\prime}, c^{\prime}$ form a four-element subset $\left\{b, c, b^{\prime}, c^{\prime}\right\}$ of $S$, which can yield a good subset $\{x, y, z, w\}$ of $S$, and at most two pairs $(b, c)$ form the same subset $\{x, y, z, w\}$ (only $(b, c)=(w, z)$ or $(w, y))$. Therefore, $S$ has at least 46 good subsets.

On the other hand, the number of good subsets $\{x, y, z, w\}$ of $S$ is equal to $\sum \frac{1}{2} s_{i}\left(s_{i}-1\right)$, where $s_{i}$ is the number of pairs $(b, c)$ in $S$ such that $b+c=i\left(i \in \mathbf{N}_{+}\right)$ and $b \leqslant c$.

Note that for each $i$, each element $s$ in $S$ appears in at most one of the pairs $(b, c)$ (in fact, when $s \leqslant i-s$, $s$ appears in the pair $(s, i-s)$, and in other cases, it appears in $(i-s, s)$), so $s_{i} \leqslant 10$. Therefore, when $s_{i} \neq 0$, $1 \leqslant s_{i} \leqslant 10$.
Thus, $\frac{1}{2} s_{i}\left(s_{i}-1\right) \leqslant 5 s_{i}-5$.
Since the set $\left.\left|a_{i}+a_{j}\right| 1 \leqslant i, j \leqslant 20\right\}$ contains 201 different elements, there are 201 positive integers $i$ such that $s_{i} \geqslant 1$. Let $T$ be the set of such $i$. Using the fact that $S$ has $\mathrm{C}_{20}^{2}$ pairs $(b, c)$ such that $b<c$, and 20 pairs $(b, c)$ such that $b=c$, we have $\sum_{i \in T} s_{i}=\mathrm{C}_{20}^{2}+20=210$. Then
$$
\begin{array}{l}
\sum_{i \in T} \frac{1}{2} s_{i}\left(s_{i}-1\right) \leqslant \sum_{i \in T}\left(5 s_{i}-5\right) \\
=5(210-201)=45 .
\end{array}
$$

This contradicts the previous conclusion that $S$ has at least 46 good subsets.

Therefore, the given set must have at least 100 different elements.