Determine all functions $f: \mathbb{Z}\to\mathbb{Z}$ satisfying the following property: For each pair of integers $m$ and $n$ (not necessarily distinct), $\mathrm{gcd}(m, n)$ divides $f(m) + f(n)$.

Note: If $n\in\mathbb{Z}$, $\mathrm{gcd}(m, n)=\mathrm{gcd}(|m|, |n|)$ and $\mathrm{gcd}(n, 0)=n$.
1. **Initial Condition Analysis:**
   - Let \( n = 0 \). The given condition becomes \(\gcd(m, 0) \mid f(m) + f(0)\).
   - Since \(\gcd(m, 0) = |m|\), we have \(|m| \mid f(m) + f(0)\).
   - This implies \(m \mid f(m) + f(0)\) for all \(m \in \mathbb{Z}\).

2. **Special Case \(m = 0\):**
   - For \(m = 0\), the condition becomes \(\gcd(0, n) \mid f(0) + f(n)\).
   - Since \(\gcd(0, n) = |n|\), we have \(|n| \mid f(0) + f(n)\).
   - This implies \(n \mid f(0) + f(n)\) for all \(n \in \mathbb{Z}\).

3. **Implication for \(f(0)\):**
   - From the above, we have \(n \mid f(0) + f(n)\) for all \(n\).
   - Let \(n = 1\). Then \(1 \mid f(0) + f(1)\), which is always true since any integer divides itself.
   - Let \(n = -1\). Then \(-1 \mid f(0) + f(-1)\), which is also always true.
   - However, for \(n = 2\), we get \(2 \mid f(0) + f(2)\). This implies \(f(0) + f(2)\) must be even.
   - Similarly, for \(n = 3\), we get \(3 \mid f(0) + f(3)\). This implies \(f(0) + f(3)\) must be divisible by 3.
   - Continuing this for all \(n\), we see that \(f(0)\) must be such that \(f(0) + f(n)\) is divisible by \(n\) for all \(n\).

4. **Conclusion for \(f(0)\):**
   - The only way \(f(0) + f(n)\) can be divisible by every integer \(n\) is if \(f(0) = 0\).
   - Therefore, \(f(0) = 0\).

5. **General Case for \(f(n)\):**
   - Given \(f(0) = 0\), the condition simplifies to \(\gcd(m, n) \mid f(m) + f(n)\).
   - For \(m = n\), we have \(\gcd(n, n) = n\), so \(n \mid 2f(n)\).
   - This implies \(n \mid f(n)\) for all \(n\).

6. **Verification:**
   - If \(f(n) = kn\) for some integer \(k\), then \(f(m) = km\) and \(f(n) = kn\).
   - The condition \(\gcd(m, n) \mid f(m) + f(n)\) becomes \(\gcd(m, n) \mid km + kn = k(m + n)\).
   - Since \(\gcd(m, n) \mid m\) and \(\gcd(m, n) \mid n\), it follows that \(\gcd(m, n) \mid k(m + n)\).
   - Therefore, \(f(n) = kn\) satisfies the condition for any integer \(k\).

The final answer is \( \boxed{ f(n) = kn } \) for any integer \(k\).