A circular track with diameter $500$ is externally tangent at a point A to a second circular track with diameter $1700.$  Two runners start at point A at the same time and run at the same speed.  The first runner runs clockwise along the smaller track while the second runner runs clockwise along the larger track.  There is a first time after they begin running when their two positions are collinear with the point A.  At that time each runner will have run a distance of $\frac{m\pi}{n}$ where  $m$  and  $n$ are relatively prime positive integers.  Find  $m + n. $
1. **Define the problem and setup the geometry:**
   - Let the diameter of the smaller track be \(500\) units, so its radius \(r_1 = \frac{500}{2} = 250\) units.
   - Let the diameter of the larger track be \(1700\) units, so its radius \(r_2 = \frac{1700}{2} = 850\) units.
   - Let the centers of the smaller and larger circles be \(D\) and \(E\) respectively.
   - Both circles are externally tangent at point \(A\).

2. **Determine the angular displacement:**
   - Let the first runner on the smaller track run a distance \(x\).
   - The angular displacement \(\theta_1\) for the first runner is given by:
     \[
     \theta_1 = \frac{x}{r_1} = \frac{x}{250}
     \]
   - Let the second runner on the larger track run a distance \(y\).
   - The angular displacement \(\theta_2\) for the second runner is given by:
     \[
     \theta_2 = \frac{y}{r_2} = \frac{y}{850}
     \]

3. **Condition for collinearity:**
   - For the runners to be collinear with point \(A\), the angles \(\theta_1\) and \(\theta_2\) must satisfy:
     \[
     \theta_1 = \theta_2 + k \cdot 2\pi \quad \text{for some integer } k
     \]
   - Since both runners start at \(A\) and run at the same speed, the distances they run are proportional to their respective circumferences. Therefore, we need:
     \[
     \frac{x}{2\pi \cdot 250} = \frac{y}{2\pi \cdot 850}
     \]
     Simplifying, we get:
     \[
     \frac{x}{500\pi} = \frac{y}{1700\pi} \implies \frac{x}{500} = \frac{y}{1700} \implies y = \frac{1700}{500}x = 3.4x
     \]

4. **Finding the first collinear point:**
   - The first time they are collinear after starting is when the smaller circle runner completes one full revolution and the larger circle runner completes a fraction of its revolution.
   - Let \(x = 500k\) where \(k\) is the number of revolutions of the smaller circle runner.
   - For the first collinear point, \(k = 1\):
     \[
     x = 500
     \]
   - Then, the distance run by the second runner is:
     \[
     y = 3.4 \times 500 = 1700
     \]

5. **Calculate the distance run by each runner:**
   - The distance run by the first runner is \(x = 500\).
   - The distance run by the second runner is \(y = 1700\).

6. **Express the distance in the form \(\frac{m\pi}{n}\):**
   - Since the distance run by each runner is proportional to their respective circumferences, we need to find the smallest \(m\) and \(n\) such that:
     \[
     \frac{500}{500\pi} = \frac{m}{n\pi} \implies m = 500, n = 1
     \]

7. **Sum of \(m\) and \(n\):**
   - \(m + n = 500 + 1 = 501\)

The final answer is \(\boxed{501}\)