, 2021

Assume the contrary. Say that the $k^{\text{th}}$ thimble is the central thimble of the $k^{\text{th}}$ move, and its position on that move is the central position of the move.

Step 1: Black and white colouring.
Before the moves start, let us paint all thimbles in white. Then, after each move, we repaint its central thimble in black. This way, at the end of the process all thimbles have become black.
By our assumption, in every move $k$, the two swapped thimbles have the same colour (as their numbers are either both larger or both smaller than $k$). At every moment, assign the colours of the thimbles to their current positions; then the only position which changes its colour in a move is its central position. In particular, each position is central for exactly one move (when it is being repainted to black).

Step 2: Red and green colouring.
Now we introduce a colouring of the positions. If in the $k^{\text{th}}$ move, the numbers of the two swapped thimbles are both less than $k$, then we paint the central position of the move in red; otherwise we paint that position in green. This way, each position has been painted in red or green exactly once. We claim that among any two adjacent positions, one becomes green and the other one becomes red; this will provide the desired contradiction since 2021 is odd.

Consider two adjacent positions $A$ and $B$, which are central in the $a^{\text{th}}$ and in the $b^{\text{th}}$ moves, respectively, with $a < b$. Then in the $a^{\text{th}}$ move the thimble at position $B$ is white, and therefore has a number greater than $a$. After the $a^{\text{th}}$ move, position $A$ is green and the thimble at position $A$ is black. By the arguments from Step 1, position $A$ contains only black thimbles after the $a^{\text{th}}$ step. Therefore, on the $b^{\text{th}}$ move, position $A$ contains a black thimble whose number is therefore less than $b$, while thimble $b$ is at position $B$. So position $B$ becomes red, and hence $A$ and $B$ have different colours.