Find all positive integers $m, n$ such that $n + (n+1) + (n+2) + ...+ (n+m) = 1000$.	
1. We start with the given equation:
   \[
   n + (n+1) + (n+2) + \cdots + (n+m) = 1000
   \]

2. This is the sum of an arithmetic series where the first term is \(n\) and the last term is \(n+m\). The number of terms in this series is \(m+1\).

3. The sum of an arithmetic series can be calculated using the formula:
   \[
   \text{Sum} = \frac{\text{number of terms}}{2} \times (\text{first term} + \text{last term})
   \]
   Here, the number of terms is \(m+1\), the first term is \(n\), and the last term is \(n+m\).

4. Substituting these values into the formula, we get:
   \[
   \frac{m+1}{2} \times (n + (n+m)) = 1000
   \]

5. Simplifying inside the parentheses:
   \[
   \frac{m+1}{2} \times (2n + m) = 1000
   \]

6. Multiplying both sides by 2 to clear the fraction:
   \[
   (m+1)(2n + m) = 2000
   \]

7. Expanding the left-hand side:
   \[
   2n(m+1) + m(m+1) = 2000
   \]

8. Rearranging the equation to solve for \(n\):
   \[
   2n(m+1) = 2000 - m(m+1)
   \]
   \[
   2n(m+1) = 2000 - m^2 - m
   \]
   \[
   2n(m+1) = 2000 - m(m+1)
   \]

9. Dividing both sides by \(2(m+1)\):
   \[
   n = \frac{2000 - m(m+1)}{2(m+1)}
   \]

10. For \(n\) to be a positive integer, \(\frac{2000 - m(m+1)}{2(m+1)}\) must be an integer. This implies that \(m+1\) must be a divisor of 2000.

11. The divisors of 2000 are: 1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 125, 200, 250, 400, 500, 1000, 2000.

12. We need to check which of these values for \(m+1\) result in \(n\) being a positive integer:
    - For \(m+1 = 5\):
      \[
      m = 4
      \]
      \[
      n = \frac{2000 - 4 \cdot 5}{2 \cdot 5} = \frac{2000 - 20}{10} = \frac{1980}{10} = 198
      \]
      This gives the solution \((m, n) = (4, 198)\).

    - For \(m+1 = 8\):
      \[
      m = 7
      \]
      \[
      n = \frac{2000 - 7 \cdot 8}{2 \cdot 8} = \frac{2000 - 56}{16} = \frac{1944}{16} = 121.5
      \]
      This does not give an integer solution.

    - For \(m+1 = 10\):
      \[
      m = 9
      \]
      \[
      n = \frac{2000 - 9 \cdot 10}{2 \cdot 10} = \frac{2000 - 90}{20} = \frac{1910}{20} = 95.5
      \]
      This does not give an integer solution.

    - For \(m+1 = 25\):
      \[
      m = 24
      \]
      \[
      n = \frac{2000 - 24 \cdot 25}{2 \cdot 25} = \frac{2000 - 600}{50} = \frac{1400}{50} = 28
      \]
      This gives the solution \((m, n) = (24, 28)\).

    - For \(m+1 = 40\):
      \[
      m = 39
      \]
      \[
      n = \frac{2000 - 39 \cdot 40}{2 \cdot 40} = \frac{2000 - 1560}{80} = \frac{440}{80} = 5.5
      \]
      This does not give an integer solution.

    - For \(m+1 = 50\):
      \[
      m = 49
      \]
      \[
      n = \frac{2000 - 49 \cdot 50}{2 \cdot 50} = \frac{2000 - 2450}{100} = \frac{-450}{100} = -4.5
      \]
      This does not give an integer solution.

    - For \(m+1 = 125\):
      \[
      m = 124
      \]
      \[
      n = \frac{2000 - 124 \cdot 125}{2 \cdot 125} = \frac{2000 - 15500}{250} = \frac{-13500}{250} = -54
      \]
      This does not give an integer solution.

    - For \(m+1 = 200\):
      \[
      m = 199
      \]
      \[
      n = \frac{2000 - 199 \cdot 200}{2 \cdot 200} = \frac{2000 - 39800}{400} = \frac{-37800}{400} = -94.5
      \]
      This does not give an integer solution.

    - For \(m+1 = 500\):
      \[
      m = 499
      \]
      \[
      n = \frac{2000 - 499 \cdot 500}{2 \cdot 500} = \frac{2000 - 249500}{1000} = \frac{-247500}{1000} = -247.5
      \]
      This does not give an integer solution.

    - For \(m+1 = 1000\):
      \[
      m = 999
      \]
      \[
      n = \frac{2000 - 999 \cdot 1000}{2 \cdot 1000} = \frac{2000 - 999000}{2000} = \frac{-997000}{2000} = -498.5
      \]
      This does not give an integer solution.

13. Therefore, the only valid solutions are:
    \[
    (m, n) = (4, 198) \quad \text{and} \quad (m, n) = (24, 28)
    \]

The final answer is \( \boxed{ (4, 198) } \) and \((24, 28)\).