Let as many odd numbers as we please, AB, BC, CD, and DE, even in multitude, be added together.
I say that the sum AE is even.
Since each of the numbers AB, BC, CD, and DE is odd, if a unit is subtracted from each, then each of the remainders is even, so that the sum of them is even. But the multitude of the units is also even. Therefore the sum AE is also even.
Therefore, if as many odd numbers as we please are added together, and their multitude is even, then the sum is even.