9. Let set $A$ consist entirely of positive integers, and for any $x, y \in A (x \neq y)$, we have $|x-y| \geqslant \frac{1}{25} x y$. How many numbers can $A$ contain at most?
9. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}\left(a_{1}<a_{2}<\cdots<a_{n}\right)$, and for any $k \ (1 \leqslant k \leqslant n)$, there exists $a_{i} \in A$ such that $a_{i} \in \left(\frac{k}{2}, \frac{k+1}{2}\right)$. Then, for any $k \ (1 \leqslant k \leqslant n)$, we have $\frac{a_{k}}{k}>\frac{n-k}{25}$, and $a_{k} \geqslant k$, so $\frac{1}{k}>\frac{n-k}{25}, n<k+\frac{25}{k}$, thus $n<\min \left\{k+\frac{25}{k}\right\}=2 \sqrt{k \cdot \frac{25}{k}}=10$. On the other hand, it is easy to verify that the set $A=\{1,2,3,4,5,7,10,16,45\}$ of 9 numbers satisfies the conditions of the problem, hence $A$ contains at most 9 numbers.