972 B
The key to solve this problem is effectively iterating through the permutations for a given width and height (perms[w, h]).
First we look at the baseline with width=1. The basic case perms[1,1] is 1.
After that perms[1,h] = perms[1,h-1] + h (so we can iterate easily through all these solutions).
With the baseline in place we can see that perms[w, h] = perms[w, h-1] + perms[w, 1] * h.
Then we just iterate through all the possibilities and search for the smallest difference.
We can stop increasing the width when perms[w, 1] > 2,000,000 and similar stop increasing the height when perms[w, h] > 2,000,000 or w > h.
The second conditions stems from the fact that perms[w, h] == perms[h, w] (it's a mirrored functions).
Through these limiting conditions and the fact that each step is pretty fast (just a few additions and multiplications) this algorithm is really fast. (We only test around 10000 values before our search space is depleted)