60th Belarusian Mathematical Olympiad

Note that if $a_1,\dots ,a_n$ satisfy the equality $$a_1+\cdots +a_n=r\left(\frac{1}{a_1}+\cdots +\frac{1}{a_n}\right),(\ast )$$ then the numbers $b_1=\frac{r}{a_1},\dots ,b_n=\frac{r}{a_n}$ satisfy the equality $$b_1+\cdots +b_n=r\left(\frac{1}{b_1}+\cdots +\frac{1}{b_n}\right).$$ By condition, we have $$\frac{1}{\sqrt{r}-b_1}+\cdots +\frac{1}{\sqrt{r}-b_n}=\frac{1}{\sqrt{r}},$$ which gives $\frac{1}{\sqrt{r}-\frac{r}{a_1}}+\cdots +\frac{1}{\sqrt{r}-\frac{r}{a_n}}=\frac{1}{\sqrt{r}}$. The last equality is equivalent to the equality $$\frac{a_1}{\sqrt{r}-a_1}+\cdots +\frac{a_n}{\sqrt{r}-a_n}=-1.(1)$$ By condition, $$\frac{1}{\sqrt{r}-a_1}+\cdots +\frac{1}{\sqrt{r}-a_n}=\frac{1}{\sqrt{r}}.(2)$$ Subtracting (1) from (2) multiplied by $\sqrt{r}$, we obtain $$\frac{\sqrt{r}-a_1}{\sqrt{r}-a_1}+\cdots +\frac{\sqrt{r}-a_n}{\sqrt{r}-a_n}=1-(-1)=2.$$ It follows that $n=2$.

It remains to verify that the problem condition holds for $n=2$. Indeed $$a_1+a_2=r\left(\frac{1}{a_1}+\frac{1}{a_2}\right)⟺(a_1+a_2)\left(1-\frac{r}{a_1{a}_2}\right)=0.$$ By condition, $a_1,a_2\ge 0$ hence the last equality is equivalent to the equality $a_1{a}_2=r$. Then $$\frac{1}{\sqrt{r}-a_1}+\frac{1}{\sqrt{r}-a_2}=\frac{2\sqrt{r}-(a_1+a_2)}{r-\sqrt{r}(a_1+a_2)+a_1{a}_2}=\frac{2\sqrt{r}-(a_1+a_2)}{2r-\sqrt{r}(a_1+a_2)}=\frac{1}{\sqrt{r}},$$ as required.