Example 11 (1) Find all natural numbers $n(n \geqslant 2)$, such that there exist real numbers $a_{1}, a_{2}, \cdots, a_{n}$, satisfying $||a_{i}-a_{j}|$ $\{1 \leqslant i<j \leqslant n\}=\left\{1,2, \cdots, \frac{n(n-1)}{2}\right\}$.
(2) Let $A=\{1,2,3,4,5,6\}, B=\{7,8,9, \cdots, n\}$. Choose three numbers from $A$ and two numbers from $B$ to form a set of five elements $A_{i}, i=1,2, \cdots, 20,\left|A_{i} \cap A_{j}\right| \leqslant 2,1 \leqslant i<j \leqslant 20$. Find the minimum value of $n$.
(2002 China National Team Training Selection Contest Problem)
(1) $a_{1}, a_{2}, \cdots, a_{n}$ have the following properties:
(i) $a_{1}, a_{2}, \cdots, a_{n}$ are all distinct;
(ii) The absolute values of their differences are all distinct.

Thus, $n=2, a_{1}=0, a_{2}=1$;
$$
\begin{array}{l}
n=3, a_{1}=0, a_{2}=1, a_{3}=3 ; \\
n=4, a_{1}=0, a_{2}=2, a_{3}=5, a_{4}=6 .
\end{array}
$$

We now prove: When $n \geqslant 5$, there do not exist $a_{1}, a_{2}, \cdots, a_{n}$ that satisfy the given conditions.
Let $0 \leqslant a_{1} < a_{2} < \cdots < a_{n}$, and $b_{i}=a_{i+1}-a_{i}, i>1$. It is known that there exists an index $i$ such that $b_{i+1}=2$, thus $b_{i}>n-2$. Therefore, $b_{i}=n-1$. This implies $b_{3}=2, b_{4}=n-2$, at this point, $b_{1}+b_{2}=b_{3}+b_{4}$, leading to a contradiction. Therefore, when $n-1 \geqslant 4$, i.e., $n \geqslant 5$, there do not exist $a_{1}, \cdots, a_{n}$ that satisfy the given conditions.

Assume $b_{n-1}=1, b_{n-2}=n-1$. The same discussion as above still leads to the conclusion that when $n \geqslant 5$, there do not exist $a_{1}, a_{2}, \cdots, a_{n}$ that satisfy the given conditions.
(2) The minimum value of $n$ is 16.
Assume each number in $B$ appears at most $k$ times in all $A_{i}$, then $k \leqslant 4$. Otherwise, if a number $m$ appears $k$ times, $k>4, 3k>12$, in the $A_{i}$ where $m$ appears, at least one $A$ has a number appearing 3 times, let's say it is 1, then there are sets $\left\{1, a_{1}, a_{2}, m, b\right\},\left\{1, a_{3}, a_{4}, m, b_{2}\right\},\left\{1, a_{5}, a_{6}, m, b_{3}\right\}$, where $a_{i} \in A, 1 \leqslant i \leqslant 6$. To satisfy the condition, $a_{i}$ must all be distinct, but there are only 5 numbers: $2,3,4,5,6$. This is impossible.
$k \leqslant 4, 20$ sets $A_{i}$, $B$ contains 40 numbers, thus there must be at least 10 different numbers, $6+10=16$, so $n \geqslant 16$. When $n=16$, the following 20 sets can be constructed:
$$
\begin{array}{l}
\{1,2,3,7,8\},\{1,2,4,12,14\},\{1,2,5,15,16\},\{1,2,6,9,10\},\{1,3,4,10,11\}, \\
\{1,3,5,13,14\},\{1,3,6,12,15\},\{1,4,5,7,9\},\{1,4,6,13,16\},\{1,5,6,8,11\}, \\
\{2,3,4,13,15\},\{2,3,5,9,11\},\{2,3,6,14,16\},\{2,4,5,8,10\},\{2,4,6,7,11\}, \\
\{2,5,6,12,13\},\{3,4,5,12,16\},\{3,4,6,8,9\},\{3,5,6,7,10\},\{4,5,6,14,15\} .
\end{array}
$$