Let $R=(8,6)$. The lines whose equations are $8y=15x$ and $10y=3x$ contain points $P$ and $Q$, respectively, such that $R$ is the midpoint of $\overline{PQ}$. The length of $PQ$ equals $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
1. **Identify the equations of the lines and the coordinates of point \( R \):**
   - The equations of the lines are \( 8y = 15x \) and \( 10y = 3x \).
   - Point \( R \) is given as \( (8, 6) \).

2. **Express the equations of the lines in slope-intercept form:**
   - For \( 8y = 15x \):
     \[
     y = \frac{15}{8}x
     \]
   - For \( 10y = 3x \):
     \[
     y = \frac{3}{10}x
     \]

3. **Assume the coordinates of points \( P \) and \( Q \) on the respective lines:**
   - Let \( P = (x_1, y_1) \) on the line \( y = \frac{15}{8}x \):
     \[
     y_1 = \frac{15}{8}x_1
     \]
   - Let \( Q = (x_2, y_2) \) on the line \( y = \frac{3}{10}x \):
     \[
     y_2 = \frac{3}{10}x_2
     \]

4. **Use the midpoint formula to find the coordinates of \( P \) and \( Q \):**
   - The midpoint \( R \) of \( \overline{PQ} \) is given by:
     \[
     R = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (8, 6)
     \]
   - This gives us two equations:
     \[
     \frac{x_1 + x_2}{2} = 8 \quad \text{and} \quad \frac{y_1 + y_2}{2} = 6
     \]

5. **Solve for \( x_1 \) and \( x_2 \):**
   - From \( \frac{x_1 + x_2}{2} = 8 \):
     \[
     x_1 + x_2 = 16
     \]

6. **Solve for \( y_1 \) and \( y_2 \):**
   - From \( \frac{y_1 + y_2}{2} = 6 \):
     \[
     y_1 + y_2 = 12
     \]

7. **Substitute \( y_1 \) and \( y_2 \) in terms of \( x_1 \) and \( x_2 \):**
   - \( y_1 = \frac{15}{8}x_1 \)
   - \( y_2 = \frac{3}{10}x_2 \)
   - Substitute these into \( y_1 + y_2 = 12 \):
     \[
     \frac{15}{8}x_1 + \frac{3}{10}x_2 = 12
     \]

8. **Solve the system of equations:**
   - We have:
     \[
     x_1 + x_2 = 16
     \]
     \[
     \frac{15}{8}x_1 + \frac{3}{10}x_2 = 12
     \]
   - Multiply the second equation by 40 to clear the denominators:
     \[
     75x_1 + 12x_2 = 480
     \]
   - Now solve the system:
     \[
     x_1 + x_2 = 16
     \]
     \[
     75x_1 + 12x_2 = 480
     \]

9. **Express \( x_2 \) in terms of \( x_1 \) and substitute:**
   - From \( x_1 + x_2 = 16 \):
     \[
     x_2 = 16 - x_1
     \]
   - Substitute into the second equation:
     \[
     75x_1 + 12(16 - x_1) = 480
     \]
     \[
     75x_1 + 192 - 12x_1 = 480
     \]
     \[
     63x_1 = 288
     \]
     \[
     x_1 = \frac{288}{63} = \frac{32}{7}
     \]
   - Then:
     \[
     x_2 = 16 - \frac{32}{7} = \frac{80}{7}
     \]

10. **Find \( y_1 \) and \( y_2 \):**
    - \( y_1 = \frac{15}{8} \cdot \frac{32}{7} = \frac{60}{7} \)
    - \( y_2 = \frac{3}{10} \cdot \frac{80}{7} = \frac{24}{7} \)

11. **Calculate the distance \( PQ \):**
    - Use the distance formula:
      \[
      PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
      \]
      \[
      PQ = \sqrt{\left( \frac{80}{7} - \frac{32}{7} \right)^2 + \left( \frac{24}{7} - \frac{60}{7} \right)^2}
      \]
      \[
      PQ = \sqrt{\left( \frac{48}{7} \right)^2 + \left( -\frac{36}{7} \right)^2}
      \]
      \[
      PQ = \sqrt{\frac{2304}{49} + \frac{1296}{49}}
      \]
      \[
      PQ = \sqrt{\frac{3600}{49}} = \frac{60}{7}
      \]

12. **Express the length \( PQ \) as \( \frac{m}{n} \) and find \( m+n \):**
    - Here, \( m = 60 \) and \( n = 7 \).
    - Therefore, \( m+n = 60 + 7 = 67 \).

The final answer is \( \boxed{67} \)