Hellenic Mathematical Olympiad

We seek solutions $(a, b)$ with $a - b = k b$, that is $a = (k+1) b$ for some integer $k > 0$. Then the left part of the given equation is equal to
$$\frac{1}{k b} - \frac{1}{(k+1) b} + \frac{1}{b} = \frac{1}{b} \left( \frac{1}{k} - \frac{1}{k+1} + 1 \right) = \frac{1}{b} \cdot \frac{k(k+1)+1}{k(k+1)}$$
and so we have
$$b = \frac{(k(k + 1) + 1)n}{k(k + 1)} = n + \frac{n}{k(k + 1)}.$$
Therefore, it is enough to select as $n$ a positive integer with more than 2024 positive integers of the form $k(k + 1)$. In fact, for example, if $(p_n)$ is the sequence of primes with $p_1=2, p_2=3, \dots$, then the number
$$n = p_1(p_1+1) p_2(p_2+1) \cdots p_{2025}(p_{2025}+1),$$
satisfies the problem.

The equation is written in the form
$$\frac{a^2 - ab + b^2}{ab(a - b)} = \frac{1}{n}.$$
We put $d = (a, b)$, and so $a = d x, b = d y$ and the equation becomes:
$$\frac{x y (x - y)}{x^2 - x y + y^2} = \frac{n}{d}$$
If $(n, d) = S$, then $n = S u$ and $d = S v$. Therefore it follows
$$x y (x - y) = u, \quad x^2 - x y + y^2 = v.$$
By putting $x y = s$ and $x - y = t$, the two relations take the form $s t = u$ and $t^2 - 3 s = v$. Then $x = \frac{s}{y}$ and $\frac{s}{y} - y = t \Leftrightarrow y^2 + y t - s = 0$.
The discriminant of the last equation is $\Delta = t^2 + 4 s$. In order to integer solution, it is enough to choose $s = k (t + k)$ for some positive integer $k$. Then the equation has the solution $y = k$.
This gives $x = k + t$, which in turn gives $u = s t = k t (k + t)$ and $v = t^2 - 3 k (t + k)$.
We should have $v > 0$ (one way to achieve this is, $k = 1$ and $t \ge 4$).
The construction now has as follows: We select a positive integer $n$ with 2024 positive integers of the form $k t (k + t)$. For each one of them is determined a triad $(x, y, v)$. From this triad is determined a pair $(a, b)$ satisfying the equation.