Remark
The proof is only true if m and n are equal. Make it more general.
m = 2k
n = 2v
m + n = 2k + 2v = 2(k + v).
k and v can be equal but many times they are not. From that simple equation you cannot do anything for sure but divide by 2.
There are 4 combinations
m is divisible by 4 and n is not. The result will not be divisible by 4.
m is not divisible by 4 but n is. The result will not be divisible by 4.
But are divisible by 4 then the sum will be as well. Here's the really odd result
If both are even and not divisible by 4 then their sum is divisible by 4
