Belarusian Mathematical Olympiad

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

It remains to verify that the problem condition holds for $n = 2$. Indeed
$$a_1 + a_2 = r \left( \frac{1}{a_1} + \frac{1}{a_2} \right) \iff (a_1 + a_2) \left( 1 - \frac{r}{a_1 a_2} \right) = 0.$$
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.