Hellenic Mathematical Olympiad

We seek solutions $(a,b)$ with $a-b=kb$, that is $a=(k+1)b$ for some integer $k>0$. Then the left part of the given equation is equal to $$\frac{1}{kb}-\frac{1}{(k+1)b}+\frac{1}{b}=\frac{1}{b}\left(\frac{1}{k}-\frac{1}{k+1}+1\right)=\frac{1}{b}\cdot \frac{k(k+1)+1}{k(k+1)}$$ 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.

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Alternative solution.

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 $a=dx,b=dy$ and the equation becomes: $$\frac{xy(x-y)}{x^2-xy+y^2}=\frac{n}{d}$$ If $(n,d)=S$, then $n=Su$ and $d=Sv$. Therefore it follows $$xy(x-y)=u,x^2-xy+y^2=v.$$ By putting $xy=s$ and $x-y=t$, the two relations take the form $st=u$ and $t^2-3s=v$. Then $x=\frac{s}{y}$ and $\frac{s}{y}-y=t\iff y^2+yt-s=0$. The discriminant of the last equation is $\Delta =t^2+4s$. 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 $u=st=kt(k+t)$ and $v=t^2-3k(t+k)$. 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 $kt(k+t)$. For each one of them is determined a triad $(x,y,v)$. From this triad is determined a pair $(a,b)$ satisfying the equation.