On the $ n $th iteration, we list $ n + 3 $ integers and skip $ n $ integers. So after $ n $ iterations, the last skipped integer is \[\sum_{k = 1}^n (2k + 3) = 2 \sum_{k = 1}^n k + 3n = n(n + 1) + 3n = n^2 + 4n.\]The number of integers we actually write down is \[\sum_{k = 1}^n (k + 3) = \sum_{k = 1}^n k + 3n = \frac{n(n + 1)}{2} + 3n = \frac{n^2 + 7n}{2}.\]So after the 996th iteration, we will have written down \[\frac{996^2 + 7 \cdot 996}{2} = 499494\]integers, so we need to write another $ 500000 - 499494 = 506 $ integers to get to the 500000th term. The last skipped integer is $ 996^2 + 4 \cdot 996 = 996000 $. At the start of the 997th iteration, we write down another 1000 integers, so the 500000th term is $ \boxed{996506} $.