Problem 1418

1. Choose a particular person from the group:
- Let's denote this person as $P$.
- $P$ corresponds with the remaining 16 people.

2. Apply the Pigeonhole Principle:
- Since there are only 3 topics, by the Pigeonhole Principle, $P$ must correspond with at least $\lceil \frac{16}{3} \rceil = 6$ people on the same topic. Let's call this topic $I$.
- Denote these 6 people as $P_1, P_2, P_3, P_4, P_5, P_6$.

3. Check for pairs within the 6 people:
- If any pair among $P_1, P_2, P_3, P_4, P_5, P_6$ corresponds on topic $I$, then $P$ and this pair form a group of three people who all correspond on topic $I$. This completes the proof.
- If no such pair exists, then all pairs among $P_1, P_2, P_3, P_4, P_5, P_6$ must correspond on topics $II$ or $III$.

4. Choose a particular person from the 6 people:
- Let's choose $P_1$ from $P_1, P_2, P_3, P_4, P_5, P_6$.
- $P_1$ corresponds with the remaining 5 people $P_2, P_3, P_4, P_5, P_6$.

5. Apply the Pigeonhole Principle again:
- Since there are only 2 topics left (topics $II$ and $III$), by the Pigeonhole Principle, $P_1$ must correspond with at least $\lceil \frac{5}{2} \rceil = 3$ people on the same topic. Let's call this topic $II$.
- Denote these 3 people as $P_2, P_3, P_4$.

6. Check for pairs within the 3 people:
- If any pair among $P_2, P_3, P_4$ corresponds on topic $II$, then $P_1$ and this pair form a group of three people who all correspond on topic $II$. This completes the proof.
- If no such pair exists, then all pairs among $P_2, P_3, P_4$ must correspond on topic $III$.

7. Conclusion:
- In this case, $P_2, P_3, P_4$ form a group of three people who all correspond on topic $III$.

Thus, in all cases, we have found at least three people who write to each other about the same topic.

$\blacksquare$