Solution:

Answer: There is exactly one function satisfying the described condition, namely $f(k)=k^{2}$ for all $k$.

For the proof, let $f$ be as required.

Step 1: Substituting $m=n=1$ gives $2 f(1)-1 \mid 2 f(1)$, hence also $2 f(1)-1 \mid 2 f(1)-(2 f(1)-1)=1$ and therefore $2 f(1)-1=1$, so $f(1)=1$.

Step 2: From now on $p$ will always stand for a prime number with $p \geq 7$. Substituting $m=n=p$ gives $2 f(p)-p^{2} \mid 2 p f(p)$ and hence also $2 f(p)-p^{2} \mid 2 p f(p)-p\left(2 f(p)-p^{2}\right)=p^{3}$, so
$$2 f(p)-p^{2} \in\left\{-p^{3},-p^{2},-p,-1,1, p, p^{2}, p^{3}\right\}$$
Since $f(p)>0$ it follows that
$$f(p) \in\left\{\frac{p^{2}-p}{2}, \frac{p^{2}-1}{2}, \frac{p^{2}+1}{2}, \frac{p^{2}+p}{2}, p^{2}, \frac{p^{3}+p^{2}}{2}\right\} .$$

Step 3: We set $m=1, n=p$ and obtain $f(p)+1-p \mid p f(p)+1$, hence also $f(p)+1-p \mid p f(p)+1-p(f(p)+1-p)=p^{2}-p+1$. Assume that $f(p) \neq p^{2}$. Then it necessarily follows (note that $p^{2}-p+1$ is odd) that $f(p)+1-p \leq 1 / 3\left(p^{2}-p+1\right)$. However, by Step 2 we have $f(p) \geq\left(p^{2}-p\right) / 2$, so it follows that
$$\begin{aligned} \frac{p^{2}-p}{2}+1-p & \leq \frac{p^{2}-p+1}{3} \\ 3 p^{2}-3 p+6-6 p & \leq 2 p^{2}-2 p+1 \\ p^{2}+5 & \leq 7 p \end{aligned}$$
which does not hold for $p \geq 7$. Hence the above assumption was false and we must have $f(p)=p^{2}$.

Step 4: Let $n \in \mathbb{N}$ be arbitrary. We set $m=p$ and obtain $f(n)+p^{2}-p n \mid p^{3}+n f(n)$, hence also $f(n)+p^{2}-p n \mid p^{3}+n f(n)-n\left(f(n)-p^{2}-p n\right)=p\left(p^{2}-p n+n^{2}\right)$. For all sufficiently large primes $p$, $f(n)$, and hence also the left-hand side of the last expression, is not divisible by $p$, therefore it follows that $f(n)+p^{2}-p n \mid p^{2}-p n+n^{2}$ and thus also $f(n)+p^{2}-p n \mid\left(f(n)+p^{2}-p n\right)-\left(p^{2}-p n+n^{2}\right)=f(n)-n^{2}$. Since the left-hand side can become arbitrarily large (there are infinitely many primes), it follows that $f(n)-n^{2}=0$ and hence $f(n)=n^{2}$.

Step 5: Checking confirms that $f(k)=k^{2}$ for all $k \in \mathbb{N}$ indeed satisfies the condition: We have $f(m)+f(n)-m n=m^{2}+n^{2}-m n \geq 2 m n-m n=m n>0$, and moreover $\left(m^{2}+n^{2}-m n\right)(m+n)=m^{3}+n^{3}=m f(m)+n f(n)$, that is, $f(m)+f(n)-m n$ is different from 0 and is a divisor of the number $m f(m)+n f(n)$.