Let $\mathbb{Z}_{>0}$ denote the set of positive integers. For any positive integer $k$, a function $f: \mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$ is called [i]$k$-good[/i] if $\gcd(f(m) + n, f(n) + m) \le k$ for all $m \neq n$. Find all $k$ such that there exists a $k$-good function.

[i]Proposed by James Rickards, Canada[/i]
1. **Understanding the Problem:**
   We need to find all \( k \) such that there exists a \( k \)-good function \( f: \mathbb{Z}_{>0} \to \mathbb{Z}_{>0} \). A function \( f \) is \( k \)-good if for all \( m \neq n \), the greatest common divisor (gcd) of \( f(m) + n \) and \( f(n) + m \) is at most \( k \).

2. **Analyzing the gcd Condition:**
   We need to ensure that \( \gcd(f(m) + n, f(n) + m) \le k \) for all \( m \neq n \). Let's consider the parity of \( m \) and \( n \):
   - If \( m \) and \( n \) have different parities, then \( f(m) + n \) and \( f(n) + m \) will have different parities if \( f(m) \) and \( f(n) \) have the same parity as \( m \) and \( n \) respectively. This ensures that \( \gcd(f(m) + n, f(n) + m) \) is even.
   - If \( m \) and \( n \) have the same parity, then \( f(m) + n \) and \( f(n) + m \) will have the same parity if \( f(m) \) and \( f(n) \) have the same parity as \( m \) and \( n \) respectively. This also ensures that \( \gcd(f(m) + n, f(n) + m) \) is even.

3. **Constructing a \( 2 \)-Good Function:**
   We need to construct a function \( f \) such that \( \gcd(f(m) + n, f(n) + m) \le 2 \) for all \( m \neq n \). We will construct this function inductively.

4. **Base Case:**
   Let \( f(1) = 2 \).

5. **Inductive Step:**
   Suppose we have constructed \( f \) for all \( m \) up to \( n-1 \). We need to define \( f(n) \) such that the gcd condition holds for all \( m < n \). Consider an integer \( C \) larger than \( f(m) + n \) for any \( m < n \). Set \( f(n) = 2^{C! + 1} - n - 1 \).

6. **Verification:**
   For any \( m < n \), we have:
   \[
   f(m) + n + f(n) + m \equiv 2^{C!+1} + 2^{c} - 2 \equiv 2 \pmod{4}
   \]
   where we assume \( f(m) = 2^{c} - m - 1 \) earlier. Thus, \( 4 \) does not divide the gcd. Also, if a prime \( p \) divides \( f(m) + n \), then \( p-1 \) divides \( C! \) and:
   \[
   f(m) + n + f(n) + m \equiv 2^{C!+1} + 2^{c} - 2 \equiv 2^{c} \not\equiv 0 \pmod{p}
   \]
   by Fermat's Little Theorem (FLT). Therefore, the gcd condition is satisfied.

7. **Conclusion:**
   The function \( f \) constructed is \( 2 \)-good. Therefore, \( k \ge 2 \) is necessary and sufficient for the existence of a \( k \)-good function.

The final answer is \( \boxed{ k \ge 2 } \).