jmc

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 $\boxed{996506}.$