Let the elements of set $A$ be positive integers, and for any $x, y \in A, x>y$, it holds that $x-y \geq \frac{x y}{25}$. What is the maximum number of elements that set $A$ can have?
I. solution. Since $x-y \geq \frac{xy}{25}$ and $y>0$, it follows that $x>\frac{xy}{25}$, from which $25>y$. Therefore, $A$ can have only one element greater than 24.

If $x$ and $y$ are two elements of the set $A$ with $x>y$, then

$$
x-y \geq \frac{xy}{25} \quad \text{implies} \quad y \leq \frac{25x}{25+x}, \quad \text{which means} \quad y \leq 25-\frac{625}{25+x}
$$

Thus, $x$ determines the maximum value of $y$, and the larger $x$ is, the larger the maximum allowed value of $y$ will be.

Therefore, to include the most elements in $A$, we should choose the largest possible values. It is therefore advisable to choose 24 as the second largest element. This determines the next largest element in the set, as the inequality $y \leq \frac{25 \cdot 24}{25+24}$ implies that the next largest element that can be included is 12, and so on, yielding the elements $8, 6, 4, 3, 2, 1$. For the largest element, the inequality $24 \leq \frac{25 \cdot x}{25+x}$ provides a lower bound, from which the minimum value for the largest number is 600. Therefore, the set $A$ can have at most 9 elements. One possible set of elements for $A$ is $\{1, 2, 3, 4, 6, 8, 12, 24, 600\}$. Of course, other 9-element sets can also be found, such as $\{1, 2, 3, 4, 5, 7, 10, 17, 54\}$.

Based on the work of Hablicsek Márton (Fazekas M. Főv. Gyak. Gimn., 10th grade) and Nagy Gábor (Kaposvár, Táncsics Mihály Gimn., 11th grade)

II. solution. If $x$ is an element of the set $A$, and the largest element smaller than $x$ is $y$, then for these, the condition $x-y \geq \frac{xy}{25}$ holds. This condition also holds for all other elements smaller than $y$, since decreasing $y$ increases the left side of the inequality, while the right side decreases.

Let's find the minimal $x$ values corresponding to the given $y$ values. The pairs $(y, x)$ are: $(1,2), (2,3), (3,4), (4,5), (5,7), (7,10), (10,17), (17,54)$. A suitable set $A$ contains the elements $\{1, 2, 3, 4, 5, 7, 10, 17, 54\}$, but other such sets can also be given, as seen in the previous solution.

We will prove that $A$ cannot have more than 9 elements.

The numbers 5 and 6 cannot both be elements of the set, since $6-5 < \frac{6 \cdot 5}{25}$.

Among the numbers 7, 8, 9, only one can be in the set $A$, since for these numbers, the value of $x-y$ is at most 2, while the minimum value of the right side is $\frac{7 \cdot 8}{25}$, which is greater than 2.

From the interval $[10,14]$, only one element can be in the set $A$, since here the value of $x-y$ is at most 4, while the minimum value of the right side is $\frac{10 \cdot 11}{25}$, which is greater than 4. Similarly, from the interval $[15,24]$, only one element can be in the set $A$, since here the value of $x-y$ is at most 9, while the minimum value of the right side is $\frac{15 \cdot 16}{25}$, which is greater than 9.

As in the previous solution, it can be shown that there cannot be two elements of $A$ that are greater than 24. Therefore, the set $A$ can indeed have at most nine elements.

Balogh János (Kaposvár, Táncsics Mihály Gimnázium, 12th grade)

Note: Many misunderstood the problem and sought the number of elements in the set of number pairs, which is of course infinite. We did not award points for these solutions. Those who did not receive full points made the mistake of presenting the 9 elements that satisfy the condition as if they were the only correct solution.