Alternate deletion around a circle of 2000

The integers $1$ to $2000$ are written clockwise around a circle. Going clockwise from $1$, you remove a number, spare the next, remove the next, spare the next, and so on until you have passed every remaining number once. Each new pass starts again at the smallest remaining number, which is removed, with removing and sparing alternating as before. Which number is the very last to be removed?

$1024$. Each pass keeps only the numbers in even positions of the current list, so after pass $k$ the survivors are multiples of $2^k$ (odd-length passes simply remove the final element too). The process ends with the largest power of $2$ not exceeding $2000$, namely $1024$.