Ways of adding $N$ nonnegative, different integers to give $mN$

Following Hao Sun's advice, let us try to turn this problem into a partition problem.

Note that finding the number of $N$ distinct, nonnegative integers that add up to $m\cdot N$ is equivalent to finding the number of partitions of $(m+1) \cdot N$ that are comprised of $N$ distinct parts. This is because there is a natural bijection between the two. Namely, to get from $N$ distinct, nonnegative integers that up to $m \cdot N$ to a partition of $(m+1) \cdot N$ with $N$ distinct parts, just add one to all of the $N$ nonnegative integers.

For instance in your example of $m = 3, N = 3$, $(6,3,0)$ would turn into $(6+1,3+1,0+1)=(7,4,1)$ which we see is a partition of $(3+1)\cdot 3 = 12$ with $3$ distinct parts.

Note that the inverse of this function is just to subtract one from all of the parts.

So now we have turned our problem into counting partitions of $(m+1)\cdot N$ with $N$ distinct parts. Well this is the same as counting partitions of $M =(m+1)\cdot N - \binom{N}{2}$ with $N$ parts. We can see this by another bijection. Namely we take our partition of $(m+1)\cdot N$ with $N$ distinct parts and subtract $N-1$ from the largest part, $N-2$ from the second largest part, ... , $1$ from the second smallest part, and $0$ from the smallest part. This will give us a partition of $M$ with $N$ parts.

For example taking $m = 3, N=3$, we saw that $(7,4,1)$ was a partition of $12$ with $3$ distinct parts. Under our map this would turn into the partition $(7-2,4-1,1-0)=(5,3,1)$ which is a partition of $(3+1) \cdot 3 - \binom{3}{2} = 9$ with $3$ parts. Note that the parts of the resulting partition need not be distinct. For instance the partition $(9,2,1)$ would get sent to $(7,1,1)$.

Therefore we just need to count how many partitions of $M$ with $N$ parts are there. This is the same as counting the number of partitions of $M$ where the largest part has size $N$. I'll leave the bijection or proof of this for you to think about.

The number of partitions of $M$ with largest part equal to $N$ is precisely the coefficient of $x^{M}$ in the generating function $$x^{N}\cdot \prod_{i=1}^{N} \frac{1}{1-x^{i}}$$ which can be found in the Wiki link of partitions that Hao Sun posted.

Mark Haiman has a nice PDF on partitions that covers part of what I detailed above: https://math.berkeley.edu/~mhaiman/math172-spring10/partitions.pdf