I would restate the problem as:
If N1 and N2 are any two odd integers where N2 ≥ N1, then (N2 - N1) is always an even integer.
An aside: Any odd integer has an even integer one away from it so any odd integer can be written as N= 2k+1 or N=2k-1 for some integer k. For example, N=17 is odd and 17 = [2(8) + 1] or [2(9) - 1].
So (N2 - N1) can be recast as [(2k + 1) - (2j +1)] for suitable k & j even or odd , and then
(N2 - N1) = [2k - 2j] = 2(k-j) which must be even for any integer choice of k & j.
That's the short proof. You can also prove it by mathematical induction, but that would take more steps.