Answer:
The proof assumes that n=m=2k, which is false in general.
Step-by-step explanation:
If n is an even number, then n=2k for some integer k. In the same way, if m is an even number, then m=2j for some integer j. It is important to write two different letters, k and j, because these integers are not necessarily equal.
For example, take n=10 and m=30. Then k=5 and j=15, so they are different. The fallacy of this proof is that it assumes k=j.
A correct proof would continue like this: by the usual laws of algebra we have that n+m=2k+2j=2(k+j). Since k+j is an integer, n+m=2p for some integer p=k+j, hence n+m is even.