There are inﬁnitely many ordered pairs $(m,n)$ of positive integers for which the sum
\[ m + ( m + 1)  +  (   m   +   2)   +... + (   n   - 1 )+n\]
is equal to the product  $mn$.  The four pairs with the smallest values of  $m$ are $(1, 1), (3, 6), (15, 35),$ and $(85, 204)$. Find three more $(m, n)$ pairs.
1. **Sum of Arithmetic Series**: The sum of the arithmetic series from \( m \) to \( n \) is given by:
   \[
   S = m + (m+1) + (m+2) + \cdots + n
   \]
   The formula for the sum of an arithmetic series is:
   \[
   S = \frac{(n - m + 1)(m + n)}{2}
   \]
   We are given that this sum is equal to the product \( mn \):
   \[
   \frac{(n - m + 1)(m + n)}{2} = mn
   \]

2. **Simplifying the Equation**: Multiply both sides by 2 to clear the fraction:
   \[
   (n - m + 1)(m + n) = 2mn
   \]
   Expand and simplify:
   \[
   n^2 + n - mn + mn - m^2 + m = 2mn
   \]
   \[
   n^2 + n - m^2 + m = 2mn
   \]
   Rearrange to form a quadratic equation in \( n \):
   \[
   n^2 + (1 - 2m)n + (m - m^2) = 0
   \]

3. **Quadratic Formula**: Solve for \( n \) using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 - 2m \), and \( c = m - m^2 \):
   \[
   n = \frac{-(1 - 2m) \pm \sqrt{(1 - 2m)^2 - 4(1)(m - m^2)}}{2(1)}
   \]
   Simplify inside the square root:
   \[
   n = \frac{2m - 1 \pm \sqrt{(2m - 1)^2 - 4(m - m^2)}}{2}
   \]
   \[
   n = \frac{2m - 1 \pm \sqrt{4m^2 - 4m + 1 - 4m + 4m^2}}{2}
   \]
   \[
   n = \frac{2m - 1 \pm \sqrt{8m^2 - 8m + 1}}{2}
   \]
   \[
   n = \frac{2m - 1 \pm \sqrt{(2m - 1)^2}}{2}
   \]
   \[
   n = \frac{2m - 1 \pm (2m - 1)}{2}
   \]

4. **Possible Values for \( n \)**: This gives us two possible values for \( n \):
   \[
   n = \frac{2m - 1 + (2m - 1)}{2} = 2m - 1
   \]
   \[
   n = \frac{2m - 1 - (2m - 1)}{2} = 0
   \]
   Since \( n \) must be a positive integer, we discard \( n = 0 \). Thus:
   \[
   n = 2m - 1
   \]

5. **Pell-like Equation**: We need to find pairs \((m, n)\) such that \( n = 2m - 1 \). This can be transformed into a Pell-like equation. Let \( a = 2m - 1 \) and \( b = 2m - 1 \). Then:
   \[
   a^2 - 2b^2 = -1
   \]
   This is a Pell-like equation of the form \( x^2 - 2y^2 = -1 \).

6. **Generating Solutions**: The solutions to this equation can be generated using the fundamental solution \((a_1, b_1) = (1, 1)\) and the recurrence relations:
   \[
   a_{k+1} = 3a_k + 4b_k
   \]
   \[
   b_{k+1} = 2a_k + 3b_k
   \]

7. **Finding More Pairs**: Using the recurrence relations, we generate the next few pairs:
   \[
   (a_2, b_2) = (3 \cdot 1 + 4 \cdot 1, 2 \cdot 1 + 3 \cdot 1) = (7, 5)
   \]
   \[
   (a_3, b_3) = (3 \cdot 7 + 4 \cdot 5, 2 \cdot 7 + 3 \cdot 5) = (41, 29)
   \]
   \[
   (a_4, b_4) = (3 \cdot 41 + 4 \cdot 29, 2 \cdot 41 + 3 \cdot 29) = (239, 169)
   \]

8. **Converting Back to \( (m, n) \) Pairs**: Convert these back to \((m, n)\) pairs:
   \[
   m = \frac{b + 1}{2}, \quad n = \frac{a + b}{2}
   \]
   For \((a_2, b_2) = (7, 5)\):
   \[
   m = \frac{5 + 1}{2} = 3, \quad n = \frac{7 + 5}{2} = 6
   \]
   For \((a_3, b_3) = (41, 29)\):
   \[
   m = \frac{29 + 1}{2} = 15, \quad n = \frac{41 + 29}{2} = 35
   \]
   For \((a_4, b_4) = (239, 169)\):
   \[
   m = \frac{169 + 1}{2} = 85, \quad n = \frac{239 + 169}{2} = 204
   \]

9. **Next Three Pairs**: Continuing the recurrence relations:
   \[
   (a_5, b_5) = (1393, 985)
   \]
   \[
   m = \frac{985 + 1}{2} = 493, \quad n = \frac{1393 + 985}{2} = 1189
   \]
   \[
   (a_6, b_6) = (8119, 5741)
   \]
   \[
   m = \frac{5741 + 1}{2} = 2871, \quad n = \frac{8119 + 5741}{2} = 6930
   \]
   \[
   (a_7, b_7) = (47321, 33461)
   \]
   \[
   m = \frac{33461 + 1}{2} = 16731, \quad n = \frac{47321 + 33461}{2} = 40391
   \]

Conclusion:
\[
(m, n) = (493, 1189), (2871, 6930), (16731, 40391)
\]

The final answer is \( \boxed{ (493, 1189), (2871, 6930), (16731, 40391) } \)