4. Given that $C$ is a fixed positive integer. Find all functions $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, such that for all positive integers $a, b$ satisfying $a+b>C$, we have
$$
(a+f(b)) \mid\left(a^{2}+b f(a)\right) .
$$
4. The function that satisfies the condition is $f(a)=k a$, where the constant $k$ is a positive integer independent of $C$.
Taking $a=1$ in equation (1) yields
$$
(1+f(b)) \mid(1+b f(1)) \text {. }
$$

This indicates that for any positive integer $b \geqslant C$, we have $f(b) \leqslant b f(1)$.
Consider a sufficiently large positive integer $n$ such that
$$
n b-f(b) \geqslant C
$$

holds for any positive integer $b$.
Taking $a=n b-f(b)$ in equation (1), we get
$$
\begin{array}{l}
n b \mid\left((n b-f(b))^{2}+b f(n b-f(b))\right) \\
\Rightarrow b \mid f^{2}(b) .
\end{array}
$$

In particular, for each prime $p$, we have $p \mid f^{2}(p)$.
$$
\text { Thus, } p \mid f(p) \text {. }
$$

Let $f(p)=k(p) p$.
Since when $p \geqslant C$, $f(p) \leqslant p f(1)$, there exist infinitely many $p$ such that $k(p)$ takes the same value $k$.

For any positive integer $a$, taking $b=p$ in equation (1), where $p$ is a sufficiently large prime and there are infinitely many such $p$ with $k(p)=k$,
$$
\begin{array}{l}
\text { then }(a+k p) \mid\left(\left(a^{2}+p f(a)\right)-a(a+k p)\right) \\
\Rightarrow(a+k p) \mid(p f(a)-p k a) .
\end{array}
$$

For sufficiently large prime $p$, since $(a+k p, p)=1$, we have $(a+k p) \mid(f(a)-k a)$.
This implies that $f(a)-k a=0$.
Therefore, $f(a)=k a$.
Upon verification, $f(a)=k a$ satisfies the condition.