Example 2 Let $A$ be a subset of the set of positive integers $\mathbf{N}^{*}$. For any $x, y \in A, x \neq y$, we have $\mid x-1$ $y \left\lvert\, \geqslant \frac{x y}{25}\right.$. Find the maximum value of $|A|$. (26th IMO Shortlist)
Let's assume $A=\left\{a_{1}\frac{n-1}{25}\right\}$. Therefore, $n-1>\frac{n-2}{25}$, so $2 \leqslant a_{2}<\frac{25}{n-2}, n<\frac{29}{2}, n \leqslant 14$. After testing, further refinement of the estimate is still needed.

Similarly, we have $3 \leqslant a_{3}<\frac{25}{n-3}, n \leqslant 11 ; 4 \leqslant a_{4}<\frac{25}{n-4}, n \leqslant 10 ; 5 \leqslant a_{5}<\frac{25}{n-5}, n \leqslant 9 ; 6 \leqslant a_{6}<\frac{25}{n-6}, n \leqslant 10$. Based on the trend of the range of $n$, we conjecture that $n \leqslant 9$ is the best estimate.

Next, we prove that there exists a 9-element subset $A$ that meets the conditions. First, when $x y \leqslant 25$, the condition $\mid x- y \left\lvert\, \geqslant \frac{x y}{25}\right.$ is obviously satisfied. Therefore, we can take $1,2,3,4,5 \in A$, at this point, $6 \notin A$. Otherwise, $|6-5|<\frac{30}{25}$, which does not meet the condition. Continuing in this manner, we find that we can take $7,10,17,54 \in A$. Therefore, $A=\{1,2,3,4,5,7,10,17,54\}$ is the desired set.

In conclusion, the maximum value of $|A|$ is 9.