Bins in a ring

Let a "step" be the process of moving one ball into the next bin. Then for any bin that you pick to try to place all the bins in, $1009$ balls must move an even number of steps and $1009$ balls must move an odd number of steps. (For example, if you pick a bin that originally had a white ball, then all the white balls must move an even number of steps clockwise and all the black balls must move an odd number of steps counterclockwise. Similarly, if you pick a bin that originally had a black ball, then all the black balls must move an even number of steps counterclockwise and all the white balls must move an odd number of steps clockwise.) Since $1009 \cdot \text{ even } + 1009 \cdot \text{ odd } = \text{ odd }$ , an odd number of total steps is required to win the game.

However, since each turn requires selecting $2$ balls and moving them, the total number of steps will be even. Therefore, it is not possible to win this game.

At each move, the parity (evenness/oddness) of the number of balls in the $1009$ bins originally holding the black balls stays the same. The game starts with $1009$ balls in these bins, and to win there would have to be an even number (either $0$ or $2018$ ) of balls in these bins. It is not possible to win.

Number all the boxes in sequence, starting with $1$ on a white ball, and ending with $2018$ on a black ball.

For all balls to be in the same box, either all the black balls must be moved to an odd box, or white to an even one.

As there are $1009$ balls of each colour, it's impossible to move all black balls into odd boxes without also moving a white ball to an even one, or vice versa (as exactly 2 balls must be moved).

Therefore it cannot be done.

Am I the only one who can't understand the statement.

Can only be won if the number of bins is a multiple of 4. Try starting with 2,4,6,8 bins etc

At each move, the parity (evenness/oddness) of the number of balls in the 1009 bins originally holding the black balls stays the same. The game starts with 1009 balls in these bins, and to win there would have to be an even number (either 0 or 2018 ) of balls in these bins. It is not possible to win.

It was more if guesswork, but if the balls are moving ing to o separate directions, it should be possible to bring all the whites to one box and all the blacks to the other, but because of the parity of 2018 you can never bring these together.

If one can see that the solution stays the same whenever there a equal odd number of both balls, then we can take the case where there is $1$ of each ball. We find that it is impossible to win in this case.

Label bins starting with white balls 0 and bins starting with black balls 1. After each move, calculate this sum across all labels: (label of the bin * number of balls in that bin). This sum starts with 1009 (an odd number), and its evenness/oddness should not change with each move. To have all 2018 balls on the same bin, this sum will need to either become 2018 or 0 (an even number), which is impossible.