Seven people of seven diﬀerent ages are attending a meeting. The seven people leave the meeting one at a
time in random order. Given that the youngest person leaves the meeting sometime before the oldest
person leaves the meeting, the probability that the third, fourth, and ﬁfth people to leave the meeting do so in order of their ages (youngest to oldest) is $\frac{m}{n}$ , where m and n are relatively prime positive integers. Find $m + n$.

1. **Determine the total number of possible orders:**
   The total number of ways the seven people can leave the meeting is \(7!\). Since the youngest person must leave before the oldest person, we divide by 2 (due to symmetry), giving us:
   \[
   \frac{7!}{2} = \frac{5040}{2} = 2520
   \]

2. **Define the people:**
   Let the youngest person be \(1\) and the oldest person be \(7\).

3. **Consider the four cases:**
   - **Case 1:** Neither \(1\) nor \(7\) are among the third, fourth, and fifth people to leave.
   - **Case 2:** \(1\) is among the third, fourth, and fifth people to leave, but \(7\) is not.
   - **Case 3:** \(7\) is among the third, fourth, and fifth people to leave, but \(1\) is not.
   - **Case 4:** Both \(1\) and \(7\) are among the third, fourth, and fifth people to leave.

4. **Calculate the number of good orders for each case:**
   - **Case 1:**
     - Choose 2 positions out of the first, second, sixth, and seventh for \(1\) and \(7\):
       \[
       \binom{4}{2} = 6
       \]
     - The remaining 5 people can be arranged in \(5!\) ways.
     - The probability that the third, fourth, and fifth people leave in order of their ages is \(\frac{1}{3!} = \frac{1}{6}\).
     - Thus, the number of good orders is:
       \[
       6 \times \frac{5!}{6} = 6 \times 120 = 120
       \]

   - **Case 2:**
     - Choose 2 positions out of the first, second, sixth, and seventh for \(7\):
       \[
       \binom{4}{2} = 6
       \]
     - \(1\) must be the third person to leave.
     - The remaining 5 people can be arranged in \(5!\) ways.
     - The probability that the fourth and fifth people leave in order of their ages is \(\frac{1}{2!} = \frac{1}{2}\).
     - Thus, the number of good orders is:
       \[
       6 \times \frac{5!}{2} = 6 \times 60 = 120
       \]

   - **Case 3:**
     - Choose 2 positions out of the first, second, sixth, and seventh for \(1\):
       \[
       \binom{4}{2} = 6
       \]
     - \(7\) must be the fifth person to leave.
     - The remaining 5 people can be arranged in \(5!\) ways.
     - The probability that the third and fourth people leave in order of their ages is \(\frac{1}{2!} = \frac{1}{2}\).
     - Thus, the number of good orders is:
       \[
       6 \times \frac{5!}{2} = 6 \times 60 = 120
       \]

   - **Case 4:**
     - \(1\) must be the third person to leave and \(7\) must be the fifth person to leave.
     - The remaining 5 people can be arranged in \(5!\) ways.
     - The number of good orders is:
       \[
       5! = 120
       \]

5. **Sum the number of good orders:**
   \[
   120 + 120 + 120 + 120 = 480
   \]

6. **Calculate the probability:**
   \[
   \frac{480}{2520} = \frac{4}{21}
   \]

7. **Find \(m + n\):**
   Since \(\frac{4}{21}\) is in simplest form, \(m = 4\) and \(n = 21\). Therefore:
   \[
   m + n = 4 + 21 = 25
   \]

The final answer is \(\boxed{25}\)