7. Let $k$ be any positive integer. If for all positive integers $m, n (m \neq n)$, we have
$$
(f(m)+n, f(n)+m) \leqslant k,
$$

then the function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$ is called “$k$-good”. Find all positive integers $k$ such that there exists a $k$-good function.
7. $k \geqslant 2$.

For any function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, let $G_{f}(m, n)=(f(m)+n, f(n)+m)$.
For any positive integer $k$, a $k$-good function is also a $(k+1)$-good function, so it suffices to prove that there does not exist a 1-good function, and there exists a 2-good function.
First, we prove: there does not exist a 1-good function.
Assume there exists a function $f$ such that for all positive integers $m, n (m \neq n)$, we have $G_{f}(m, n)=1$. If there exist two different even numbers $m, n$ such that $f(m), f(n)$ are both even, then $2 \mid G_{f}(m, n)$, a contradiction. Similarly, if there exist two different odd numbers $m, n$ such that $f(m), f(n)$ are both odd, then $2 \mid G_{f}(m, n)$, a contradiction. Otherwise, there exist an even number $m$ and an odd number $n$ such that $f(m)$ is odd and $f(n)$ is even, then $2 \mid G_{f}(m, n)$, a contradiction.
Next, we construct a 2-good function.
Define $f(n)=2^{g(n)+1}-n-1$, where the function $g$ satisfies
$$
g(1)=1, g(n+1)=\left(2^{g(n)+1}\right)!\left(n \in \mathbf{Z}_{+}\right) .
$$

For any positive integers $m, n (m>n)$, let
$$
\begin{array}{l}
A=f(m)+n=2^{g(m)+1}-m+n-1, \\
B=f(n)+m=2^{g(n)+1}-n+m-1 .
\end{array}
$$

We need to prove: $(A, B) \leqslant 2$.
Since $A+B=2^{g(m)+1}+2^{g(n)+1}-2$ is not divisible by 4, we know $4 \nmid (A, B)$. If there exists an odd prime $p$ such that $p \mid (A, B)$, the next goal is to derive a contradiction.
For each positive integer $k$, since $g(k+1)>g(k)$, then $2^{g(k+1)+1} \geqslant 2^{g(k)+1}+1$.
Repeatedly applying this inequality, we get
$$
2^{g(m-1)+1} \geqslant 2^{g(n)+1}+(m-1)-n=B \text {. }
$$

Since $p \mid B$, we have
$$
\begin{array}{l}
p-1<B \leqslant 2^{g(m-1)+1} \\
\Rightarrow(p-1) \mid\left(2^{g(m-1)+1}\right)! \\
\Rightarrow(p-1) \mid g(m) .
\end{array}
$$

By Fermat's Little Theorem, we know
$$
\begin{array}{l}
2^{g(m)} \equiv 1(\bmod p) \\
\Rightarrow A+B \equiv 2^{g(n)+1}(\bmod p) .
\end{array}
$$

Since $p \mid (A+B) \Rightarrow p=2$, a contradiction.