Let $r$ be a fixed positive real number. It is known that for some positive integer $n$ the following statement is true: for any positive real numbers $a_1,...,a_n$ satisfying the equation $a_1+...+a_n=r(\frac{1}{a_1}+...+\frac{1}{a_n})$ they also satisfy the equation $\frac{1}{\sqrt{r}-a_1}+...+\frac{1}{\sqrt{r}-a_n}=\frac{1}{\sqrt{r}}$ ($a_i\neq\sqrt{r}$). Find $n$.
1. Let \( r \) be a fixed positive real number and \( n \) be a positive integer such that for any positive real numbers \( a_1, a_2, \ldots, a_n \) satisfying the equation 
   \[
   a_1 + a_2 + \cdots + a_n = r \left( \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} \right),
   \]
   they also satisfy the equation 
   \[
   \frac{1}{\sqrt{r} - a_1} + \frac{1}{\sqrt{r} - a_2} + \cdots + \frac{1}{\sqrt{r} - a_n} = \frac{1}{\sqrt{r}}.
   \]

2. Let \( b_i = \frac{a_i}{\sqrt{r}} \). Then the given equation becomes 
   \[
   b_1 + b_2 + \cdots + b_n = \frac{1}{b_1} + \frac{1}{b_2} + \cdots + \frac{1}{b_n}.
   \]

3. We need to show that 
   \[
   \frac{1}{1 - b_1} + \frac{1}{1 - b_2} + \cdots + \frac{1}{1 - b_n} = 1.
   \]

4. Let \( \frac{1}{1 - b_i} = c_i \). Then \( b_i = 1 - \frac{1}{c_i} \) and \( \frac{1}{b_i} = 1 + \frac{1}{c_i - 1} \).

5. Substitute \( b_i \) and \( \frac{1}{b_i} \) into the equation:
   \[
   b_1 + b_2 + \cdots + b_n = \frac{1}{b_1} + \frac{1}{b_2} + \cdots + \frac{1}{b_n}
   \]
   becomes
   \[
   \left( 1 - \frac{1}{c_1} \right) + \left( 1 - \frac{1}{c_2} \right) + \cdots + \left( 1 - \frac{1}{c_n} \right) = \left( 1 + \frac{1}{c_1 - 1} \right) + \left( 1 + \frac{1}{c_2 - 1} \right) + \cdots + \left( 1 + \frac{1}{c_n - 1} \right).
   \]

6. Simplify the equation:
   \[
   n - \left( \frac{1}{c_1} + \frac{1}{c_2} + \cdots + \frac{1}{c_n} \right) = n + \left( \frac{1}{c_1 - 1} + \frac{1}{c_2 - 1} + \cdots + \frac{1}{c_n - 1} \right).
   \]

7. This implies:
   \[
   - \left( \frac{1}{c_1} + \frac{1}{c_2} + \cdots + \frac{1}{c_n} \right) = \frac{1}{c_1 - 1} + \frac{1}{c_2 - 1} + \cdots + \frac{1}{c_n - 1}.
   \]

8. Let \( d_i = 1 - c_i \). Then \( d_i - 1 = -c_i \).

9. Substitute \( d_i \) into the equation:
   \[
   \left( \frac{1}{d_1 - 1} + \frac{1}{d_1} \right) + \left( \frac{1}{d_2 - 1} + \frac{1}{d_2} \right) + \cdots + \left( \frac{1}{d_n - 1} + \frac{1}{d_n} \right) = 0.
   \]

10. This simplifies to:
    \[
    - \left( \frac{1}{c_1 - 1} + \frac{1}{c_1} + \frac{1}{c_2 - 1} + \frac{1}{c_2} + \cdots + \frac{1}{c_n - 1} + \frac{1}{c_n} \right) = 0.
    \]

11. By assertion (2), we have:
    \[
    d_1 + d_2 + \cdots + d_n = 1.
    \]

12. Therefore:
    \[
    1 - c_1 + 1 - c_2 + \cdots + 1 - c_n = 1.
    \]

13. This implies:
    \[
    c_1 + c_2 + \cdots + c_n = n - 1.
    \]

14. But by assertion (2), we have:
    \[
    c_1 + c_2 + \cdots + c_n = 1.
    \]

15. Therefore, we conclude:
    \[
    n - 1 = 1 \implies n = 2.
    \]

The final answer is \( \boxed{ n = 2 } \).