From Naturals To Naturals!

Setting $m=n$ gives: $n^2+f(n)\mid nf(n)+n~~~\Longrightarrow ~ n^2+f(n)\mid n(n^2+f(n)) - (nf(n)+n)=n^3-n$ . Setting $n=2$ gives: $f(2)+4\mid 2^3-2=6$ . Hence $f(2)=2$ . Setting $m=2, n=1$ in the original expression gives: $f(1)+4\mid 2f(2)+1$ , that is, $f(1)+4\mid 5$ . Hence $f(1)=1$ .

Now setting $m=1$ in the original expression gives: $f(n)+1\mid n+1 ~~~\Longrightarrow ~ f(n)+1\le n+1$ , that is, $f(n)\le n$ . On the other hand, setting $n=1$ in the original expression gives: $m^2+1\mid mf(m)+1~~~\Longrightarrow ~ m^2+1\le mf(m)+1$ , that is, $f(m)\geq m$ since $m$ is positive. Combining these two results, we deduce that $\boxed{f(n)=n}$ for all $n\in\mathbb{N}$ .

Therefore there's only one possible value $f(2014)=2014\equiv N$ . Hence the answer is: $2014\bmod{1000}=\fbox{14}$

Let $m = 0$ and we get $f(n) | n$ .

Now let $n = 0$ and we get $(m^2 + f(0)) | m f(m)$ . But by the previous deduction, we see that $m f(m) | m^2$ . Therefore, by transitivity, $(m^2 + f(0)) | m^2$ .

Therefore $m^2 + f(0) \le m^2$ , so $f(0) \le 0$ . On the other hand, we must have $f(0) \ge 0$ , so therefore $f(0) = 0$ .

Therefore, $m^2 | m f(m)$ , so $m | f(m)$ .

So, for all $n \in \mathbb{N}$ , we have $f(n) | n$ and $n | f(n)$ . Since the function is nonnegative, it must be that $f(n) = n$ .

Hence $f(2014) = 2014$ , so $N \pmod{1000} = \boxed{14}$ .