Solution:

Number the original positions $1$ (top) through $2000$ (bottom) and put them in a queue. Each step removes the front position (which receives the next label $1, 2, 3, \ldots$) and sends the new front to the back. So the card labelled $1999$ is the next-to-last card removed, and we must find which original position survives that long.

The first pass removes the odd positions $1, 3, \ldots, 1999$ (labels $1$ through $1000$) and, since it ends by sending $2000$ to the back, the next pass again starts by removing the front of the queue $2, 4, \ldots, 2000.$ Successive passes therefore remove $2, 6, \ldots, 1998$ (the positions $\equiv 2 \bmod 4$), then $4, 12, \ldots, 1996,$ then $8, 24, \ldots, 1992,$ then $16, 48, \ldots, 1968, 2000$ (the $63$ positions $\equiv 16 \bmod 32$). That last pass ran through an odd number ($125$) of cards, so the alternation shifts: the $62$ surviving multiples of $32$ now sit in the queue as $64, 96, \ldots, 1984, 32.$

Continuing the same removal pattern from that queue, the next rounds remove $64, 128, \ldots, 1984;$ then $96, 224, \ldots, 1888, 32;$ then $288, 544, \ldots, 1824;$ then $160, 672, 1184, 1696;$ then $416, 1440;$ and the final two cards removed are $928$ and $1952.$ So label $1999$ goes to the card at original position $928,$ which had $927$ cards above it.