Given two positive integers $m$ and $n$, find the largest $k$ in terms of $m$ and $n$ such that the following condition holds: 

Any tree graph $G$ with $k$ vertices has two (possibly equal) vertices $u$ and $v$ such that for any other vertex $w$ in $G$, either there is a path of length at most $m$ from $u$ to $w$, or there is a path of length at most $n$ from $v$ to $w$.

[i]Proposed by Ivan Chan[/i]
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 \( \boxed{\min(2n + 2m + 2, 3n + 2)} \).