Let $r$ be a positive real number. Denote by $[r]$ the integer part of $r$ and by $\{r\}$ the fractional part of $r$. For example, if $r=32.86$, then $\{r\}=0.86$ and $[r]=32$. What is the sum of all positive numbers $r$ satisfying $25\{r\}+[r]=125$?
1. Let \( r \) be a positive real number such that \( 25\{r\} + [r] = 125 \). We denote the integer part of \( r \) by \( [r] \) and the fractional part of \( r \) by \( \{r\} \).

2. Since \( [r] \) is a positive integer, let \( [r] = k \) where \( k \) is a positive integer. Then, \( 25\{r\} \) must also be an integer because \( 25\{r\} = 125 - k \).

3. Let \( 25\{r\} = m \) where \( m \) is an integer. Then, \( \{r\} = \frac{m}{25} \).

4. Substituting \( \{r\} \) back into the equation, we get:
   \[
   25 \left( \frac{m}{25} \right) + k = 125 \implies m + k = 125
   \]

5. Since \( \{r\} \) is the fractional part, \( 0 \leq \{r\} < 1 \) implies \( 0 \leq \frac{m}{25} < 1 \). Therefore, \( 0 \leq m < 25 \).

6. For each integer \( m \) from 0 to 24, there is a corresponding integer \( k = 125 - m \). Thus, there are 25 possible pairs \((k, m)\).

7. The corresponding values of \( r \) are:
   \[
   r = k + \{r\} = k + \frac{m}{25} = (125 - m) + \frac{m}{25}
   \]

8. We need to find the sum of all such \( r \):
   \[
   \sum_{m=0}^{24} \left( 125 - m + \frac{m}{25} \right)
   \]

9. Simplify the sum:
   \[
   \sum_{m=0}^{24} \left( 125 - m + \frac{m}{25} \right) = \sum_{m=0}^{24} 125 - \sum_{m=0}^{24} m + \sum_{m=0}^{24} \frac{m}{25}
   \]

10. Calculate each part separately:
    - \(\sum_{m=0}^{24} 125 = 125 \times 25 = 3125\)
    - \(\sum_{m=0}^{24} m = \frac{24 \times 25}{2} = 300\)
    - \(\sum_{m=0}^{24} \frac{m}{25} = \frac{1}{25} \sum_{m=0}^{24} m = \frac{1}{25} \times 300 = 12\)

11. Combine the results:
    \[
    3125 - 300 + 12 = 2837
    \]

The final answer is \(\boxed{2837}\).