Let $O$ be a circle with diameter $AB = 2$. Circles $O_1$ and $O_2$ have centers on $\overline{AB}$ such that $O$ is tangent to $O_1$ at $A$ and to $O_2$ at $B$, and $O_1$ and $O_2$ are externally tangent to each other. The minimum possible value of the sum of the areas of $O_1$ and $O_2$ can be written in the form $\frac{m\pi}{n}$ where $m$ and $n$ are relatively prime positive integers. Compute $m + n$.
1. **Define the problem and given values:**
   - Circle \( O \) has a diameter \( AB = 2 \).
   - Circles \( O_1 \) and \( O_2 \) have centers on \( \overline{AB} \).
   - Circle \( O \) is tangent to \( O_1 \) at \( A \) and to \( O_2 \) at \( B \).
   - Circles \( O_1 \) and \( O_2 \) are externally tangent to each other.

2. **Determine the radii of circles \( O_1 \) and \( O_2 \):**
   - Let the radius of \( O_1 \) be \( r_1 \) and the radius of \( O_2 \) be \( r_2 \).
   - Since \( O \) is tangent to \( O_1 \) at \( A \), the center of \( O_1 \) is at \( A + r_1 \).
   - Similarly, since \( O \) is tangent to \( O_2 \) at \( B \), the center of \( O_2 \) is at \( B - r_2 \).

3. **Use the condition that \( O_1 \) and \( O_2 \) are externally tangent:**
   - The distance between the centers of \( O_1 \) and \( O_2 \) is \( r_1 + r_2 \).
   - The distance between \( A + r_1 \) and \( B - r_2 \) is \( 2 - r_1 - r_2 \).

4. **Set up the equation for the external tangency:**
   \[
   r_1 + r_2 = 2 - r_1 - r_2
   \]
   Simplifying, we get:
   \[
   2(r_1 + r_2) = 2
   \]
   \[
   r_1 + r_2 = 1
   \]

5. **Minimize the sum of the areas of \( O_1 \) and \( O_2 \):**
   - The area of \( O_1 \) is \( \pi r_1^2 \).
   - The area of \( O_2 \) is \( \pi r_2^2 \).
   - We need to minimize \( \pi r_1^2 + \pi r_2^2 \).

6. **Use the constraint \( r_1 + r_2 = 1 \):**
   - Let \( r_1 = x \) and \( r_2 = 1 - x \).
   - The sum of the areas becomes:
     \[
     \pi x^2 + \pi (1 - x)^2
     \]
   - Simplify the expression:
     \[
     \pi x^2 + \pi (1 - 2x + x^2) = \pi (2x^2 - 2x + 1)
     \]

7. **Find the minimum value of the quadratic expression:**
   - The quadratic \( 2x^2 - 2x + 1 \) has its minimum value at \( x = \frac{-b}{2a} = \frac{1}{2} \).
   - Substitute \( x = \frac{1}{2} \):
     \[
     2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) + 1 = 2\left(\frac{1}{4}\right) - 1 + 1 = \frac{1}{2}
     \]
   - Therefore, the minimum sum of the areas is:
     \[
     \pi \cdot \frac{1}{2} = \frac{\pi}{2}
     \]

8. **Determine \( m \) and \( n \):**
   - The minimum possible value of the sum of the areas is \( \frac{\pi}{2} \), where \( m = 1 \) and \( n = 2 \).
   - Thus, \( m + n = 1 + 2 = 3 \).

The final answer is \( \boxed{3} \).