Given an integer $n>1$, let $a_{1}, a_{2}, \cdots, a_{n}$ be distinct non-negative real numbers, and define the sets
$$
A=\left\{a_{i}+a_{j} \mid 1 \leqslant i \leqslant j \leqslant n\right\}, B=\left\{a_{i} a_{j} \mid 1 \leqslant i \leqslant j \leqslant n\right\} .
$$

Find the minimum value of $\frac{|A|}{|B|}$. Here, $|X|$ denotes the number of elements in the set $X$.
Solution: Without loss of generality, let $a_{1}<a_{2}<\cdots<a_{n}$, then
$$
2 a_{1}<a_{1}+a_{2}<2 a_{2}<a_{2}+a_{3}<\cdots<a_{n-1}+a_{n}<2 a_{n},
$$

Therefore, $|A| \geqslant 2 n-1$ 10 points

Also, $B=\left\{a_{i}^{2} \mid i=1,2, \cdots, n\right\} \cup\left\{a_{i} a_{j} \mid 1 \leqslant i<j \leqslant n\right\}$, so
$$
|B| \leqslant n+C_{n}^{2}=\frac{n(n+1)}{2},
$$

Hence, $\frac{|A|}{|B|} \geqslant \frac{2 n-1}{\frac{n(n+1)}{2}}=\frac{2(2 n-1)}{n(n+1)}$. 20 points

When $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}=\left\{n^{2}+1, n^{2}+2, \cdots, n^{2}+n\right\}$, the equality holds.
In fact, $2 n^{2}+2 \leqslant a_{i}+a_{j} \leqslant 2 n^{2}+2 n(1 \leqslant i \leqslant j \leqslant n)$, and all integer values between $2 n^{2}+2$ and $2 n^{2}+2 n$ (including these two numbers) can be obtained, so $|A|=2 n-1$ 30 points
Furthermore, if $\left(n^{2}+i\right)\left(n^{2}+j\right)=\left(n^{2}+k\right)\left(n^{2}+l\right)$, where $i, j, k, l \in\{1,2, \cdots, n\}$. Then
$$
(i+j-k-l) n^{2}=k l-i j \text {. }
$$

Since $|k l-i j|<n^{2}$, it follows that $i+j-k-l=0$, and $k l-i j=0$, thus $\{i, j\}=\{k, l\}$, at this point $|B|=\frac{n(n+1)}{2}$.
Therefore, the minimum value of $\frac{|A|}{|B|}$ is $\frac{2(2 n-1)}{n(n+1)}$ 40 points