It depends on what the loops do. For example, if I change the end of the 2nd loop to just do 3 iterations like this:
for (int i = 0; i < N; i++)
for (int j = i; j < i+3; j++)
if (a[i] + a[j] == 0)
cnt++;
we get back to O(N)
The key is whether the number of iterations in the loop is related to N and increases linearly as N does.
Here is another example where the 2nd loop goes to N ^ 2:
for (int i = 0; i < N; i++)
for (int j = i; j < N*N; j++)
if (a[i] + a[j] == 0)
cnt++;
This would be o(N^3)