6. Find all functions $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, such that for all positive integers $m, n$, $f(m)+f(n)-m n$ is non-zero and divides $m f(m)+n f(n)$.
6. $f(n)=n^{2}\left(n \in \mathbf{Z}_{+}\right)$.

From the given condition, we have
$$
(f(m)+f(n)-m n) \mid(m f(m)+n f(n)) \text {. }
$$

Taking $m=n=1$ in equation (1), we get
$$
\begin{array}{l}
(2 f(1)-1) \mid 2 f(1) \\
\Rightarrow(2 f(1)-1) \mid(2 f(1)-(2 f(1)-1)) \\
\Rightarrow(2 f(1)-1) \mid 1 \\
\Rightarrow f(1)=1 .
\end{array}
$$

Let $p \geqslant 7$ be a prime number.
Taking $m=p, n=1$ in equation (1).
Then $(f(p)-p+1) \mid(p f(p)+1)$
$\Rightarrow(f(p)-p+1) \mid(p f(p)+1-p(f(p)-p+1))$
$\Rightarrow(f(p)-p+1) \mid\left(p^{2}-p+1\right)$.
If $f(p)-p+1=p^{2}-p+1$, then $f(p)=p^{2}$.
If $f(p)-p+1 \neq p^{2}-p+1$, since $p^{2}-p+1$ is a positive odd number, then
$$
\begin{array}{c}
p^{2}-p+1 \geqslant 3(f(p)-p+1) \\
\Rightarrow f(p) \leqslant \frac{1}{3}\left(p^{2}+2 p-2\right) .
\end{array}
$$

Taking $m=n=p$ in equation (1).
Then $\left(2 f(p)-p^{2}\right) \mid 2 p f(p)$
$\Rightarrow\left(2 f(p)-p^{2}\right) \mid\left(2 p f(p)-p\left(2 f(p)-p^{2}\right)\right)$
$\Rightarrow\left(2 f(p)-p^{2}\right) \mid p^{3}$.
From equation (2) and $f(p) \geqslant 1$, we get
$$
\begin{array}{l}
-p^{2}<2 f(p)-p^{2} \\
\leqslant \frac{2}{3}\left(p^{2}+2 p-2\right)-p^{2}<-p .
\end{array}
$$

This contradicts that $2 f(p)-p^{2}$ is a divisor of $p^{3}$.
Therefore, for all prime numbers $p \geqslant 7$, we have
$$
f(p)=p^{2} \text {. }
$$

Let $n$ be a fixed positive integer, choose a sufficiently large prime number $p$, take $m=p$ in equation (1), we get
$$
(f(p)+f(n)-p n) \mid(p f(p)+n f(n)) \text {. }
$$

Then $p f(p)+n f(n)-n(f(p)+f(n)-p n)$ is divisible by $f(p)+f(n)-p n$, i.e.,
$$
(f(p)+f(n)-p n) \mid\left(p f(p)-n f(p)+p n^{2}\right) .
$$

Since $f(p)=p^{2}$, we have
$$
\left(p^{2}-p n+f(n)\right) \mid p\left(p^{2}-p n+n^{2}\right) \text {. }
$$

Since $p$ is sufficiently large and $n$ is a fixed positive integer, $p \nmid f(n)$. This implies,
$$
\begin{array}{l}
\left(p, p^{2}-p n+f(n)\right)=1 . \\
\text { Then }\left(p^{2}-p n+f(n)\right) \mid\left(p^{2}-p n+n^{2}\right) . \\
\text { Hence }\left(p^{2}-p n+f(n)\right) \mid\left(n^{2}-f(n)\right) .
\end{array}
$$

Since $n^{2}-f(n)$ is fixed and $p^{2}-p n+f(n)$ can be sufficiently large, we must have $n^{2}-f(n)=0$, i.e., for any positive integer $n$, $f(n)=n^{2}$.
If for any positive integer $n$, $f(n)=n^{2}$, then
$$
\begin{array}{l}
f(m)+f(n)-m n=m^{2}+n^{2}-m n, \\
m f(m)+n f(n)=m^{3}+n^{3} \\
=(m+n)\left(m^{2}+n^{2}-m n\right) .
\end{array}
$$

Thus, for any positive integers $m, n$, equation (1) holds. Therefore, for any positive integer $n, f(n)=n^{2}$.