Problem 1379

1. **Assume $n \ge m$**: Without loss of generality, we can assume $n \ge m$ to simplify the problem.
2. **Determine the largest $k$**: We need to find the largest $k$ such that any tree graph $G$ with $k$ vertices has two vertices $u$ and $v$ satisfying the given condition.
3. Upper Bound Analysis:
- For $k \ge 2n + 2m + 3$, consider a path of length $2n + 2m + 3$. This path serves as a counterexample because there will be vertices that are too far apart to satisfy the condition.
- For $k \ge 3n + 3$, consider three disjoint paths with lengths $n+1, n+1, n$ respectively, and connect these paths with a new vertex at their endpoints. This construction also serves as a counterexample.
4. Lower Bound Analysis:
- Assume $k \le \min(2n + 2m + 2, 3n + 2)$ and suppose we cannot find such vertices $u$ and $v$.
- Let $A_1A_2 \dots A_t$ be a diameter of the tree, and let $G_i$ be the subtree of $A_i$ not intersecting with the diameter.
- If $t \le 2n$, choosing $v = A_{\lfloor \frac{t}{2} \rfloor}$ works because the diameter is short enough.
- Assume $t > 2n$.
5. Choosing Vertices:
- Choose $v = A_{n+1}$ and $u = A_{t-m}$. There must exist a vertex $a$ such that $d(a, v) > n$ and $d(a, u) > m$. If $a \in G_i$, then $n+1 < i < t-m$ by the diameter assumption.
- Choose $v = A_{t-n}$ and $u = A_{m+1}$. There must exist a vertex $b$ such that $d(b, v) > n$ and $d(b, u) > m$. If $b \in G_j$, then $m+1 < j < t-n$ by the diameter assumption.
6. Contradiction:
- If $i = j$, then $n+1 < i < t-n$. This implies there are at least $n+n+d(A_{n+1}, a) + 2 \ge 3n + 3$ vertices, which contradicts our assumption.
- If $i \neq j$, let $x = d(a, A_i)$ and $y = d(b, A_j)$. Summing the inequalities $x + d(A_i, A_{n+1}) \ge n+1$ and $x + d(A_{t-m}, A_i) \ge m+1$, we get $2x + (t-m-n-1) \ge m+n+2$. Similarly, we get $2y + (t-m-n-1) \ge m+n+2$. Summing these, we get $x + y + t \ge 2n + 2m + 3$. Since $i \neq j$, there are at least $x + y + t$ vertices in the tree, which is a contradiction.

Therefore, the largest $k$ is $\min(2n + 2m + 2, 3n + 2)$.

The final answer is $\boxed{\min(2n + 2m + 2, 3n + 2)}$.