20 Let $a_{1}, a_{2}, \cdots, a_{20}$ be 20 distinct positive integers, and the set $\left\{a_{i}+a_{j} \mid 1 \leqslant i \leqslant j \leqslant 20\right\}$ has 201 distinct elements. Find the minimum possible number of distinct elements in the set $\left\{\mid a_{i}-a_{j} \| 1 \leqslant i<j \leqslant 20\right\}$.
The minimum number of elements in the given set is 100.
For example, let $a_{i}=10^{11}+10^{i}, a_{10+i}=10^{11}-10^{i}, i=1$, $2, \cdots, 10$. 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, and $\left\{\left|a_{i}-a_{j}\right| \mid 1 \leqslant i \leqslant j \leqslant 20\right\}$ contains 190 pairs $(b, c)$, considering their differences $b-c$. Since there are at most 99 different differences (using the proof by contradiction assumption), 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 4-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 $(b, c)$ can form the same subset $\{x, y, z, w\}$ (only if $(b, c)=(w, z)$ and $(w, y)$). Therefore, $S$ has at least 46 "good subsets".

On the other hand, the number of "good subsets" $\{x, y, z, w\}$ of $S$ equals $\sum \frac{1}{2} s_{i} (s_{i}-1)$, where $s_{i}$ is the number of pairs $(b, c)$ in $S$ such that $b+c=i$ and $b \leqslant c$, and $i$ is a positive integer.

Notice 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 $s_{i} \neq 0$, $1 \leqslant s_{i} \leqslant 10$, and hence $\frac{1}{2} s_{i} (s_{i}-1) \leqslant 5 s_{i} - 5$. Since the set $\left\{a_{i}+a_{j} \mid 1 \leqslant i \leqslant j \leqslant 20\right\}$ 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 $\mathrm{C}_{10}^{2}$ pairs $(b, c)$ such that $b<c$ and 20 pairs $(b, c)$ such that $b=c$, we have $\sum_{i \in T} s_{i}=\mathrm{C}_{10}^{2}+20=210$. Thus,
$$
\sum_{i \in T} \frac{1}{2} s_{i} (s_{i}-1) \leqslant \sum_{i \in T} (5 s_{i} - 5) = 5 \times (210 - 201).
$$

This contradicts the previous conclusion that $S$ has at least 46 "good subsets".
Therefore, the given set must have at least 100 different elements.