Given even numbers $ a $ and $ b $, then $ a + b $ will be even as well.
Since every even number is a multiple of two, you can write $ a = 2n $ and $ b = 2m $ where $ n, m \in \Z $.
When you add $ a $ and $ b $, you can factor out a $ 2 $.
$$ a + b = 2n + 2m = 2(n + m) $$
Since $ n + m $ will always be an integer, substitute $ n + m = k, k \in \Z $.
$$ a + b = 2k $$
From this follows that $ 2k $ is always even, and thus $ a + b $ is always even as well.