The 2020 members of the society of game theorists are holding the annual election for their leadership board. All members are initially on the board, and are ranked based on their qualifications. They start off by voting on whether to keep the board the same size; if they fail to get a strict majority, the member with the lowest rank is expelled. This process continues until they finally vote to keep the size of the board intact.

It is common knowledge that, as each board member seeks to maximize their own influence, they seek to remain on the board while retaining as few other members as possible.

At the end of this process, how many society members will remain on the leadership board?
Number members \#1 to \#2020, with \#1 being the member with the highest rank. We claim that only if there are $2^{n}-1$ members still on the board for some $n \in \mathbb{N}$, they all remain on the board.

Proof. We use induction on $n$.

Base case: $n=1$.

If there were $2^{1}-1=1$ persons on the board, that person would vote to keep themselves on the board and win.

Inductive step: $n \Longrightarrow n+1$

If there were at least $2^{n}$ members, the first $2^{n}-1$ would all vote "no", because by induction they know they would be guaranteed a spot on the board in any case and want to minimize its total size. This would be a weak majority vote for "no" if the total size of the board were at most $2\left(2^{n}-1\right)=2^{n+1}-2$ members.

However, if there were $2^{n+1}-1$ members, all whose numbers are at least $2^{n}$ would vote "yes", as by the the above, they would otherwise get kicked off the board. This is a strict majority, as desired.

The largest number at most 2020 of the form $2^{n}-1$ is 1023 , so this is our answer.