Does anyone have a solution for $2023$? The solution for $2022$ is pretty straightforward: make pairs of two consecutive plates and place one coin from one plate to the plate who is also in its pair causing the sum of total pair to be an invariant of $2$ and being never more than $3$.

Am I doing something wrong? Now, if Alice picks any points that has $0$ in the left, Bob adds to the left ( reversing Alice's steps) If alice picks any points that has $2$ in the right, Bob adds to the left Otherwise, add to the right The case where $1$ is neighbour to $2$ 2 points is impossible since if we want to make $..212...$ for any $3$ consecuitive points, we must move $1$ from some $1$s and $0$ to make $2$, but then in the left there will be $0$, hence we can't add to the right. This goes infinitely many times, hence conclusion ( which makes all $n$ possible?) I think I am sillying hard Remark: By "left" and "right" , I meant counter-clockwise neighbour and clockwise neighbour of the point.

Iora wrote: Am I doing something wrong? Now, if Alice picks any points that has $0$ in the left, Bob adds to the left ( reversing Alice's steps) If alice picks any points that has $2$ in the right, Bob adds to the left Otherwise, add to the right The case where $1$ is neighbour to $2$ 2 points is impossible since if we want to make $..212...$ for any $3$ consecuitive points, we must move $1$ from some $1$s and $0$ to make $2$, but then in the left there will be $0$, hence we can't add to the right. This goes infinitely many times, hence conclusion ( which makes all $n$ possible?) I think I am sillying hard Okay so I think your idea is right, only it isn't clear what you mean with 'in the left'. Does this have to be a neighbouring plate? Call a plate with $0$ or $2$ coins 'straight'. Call two plates containing $2$ coins 'coupled', if the only coins between them are plates with more than or equal $1$ coin. Anyway completing your reasoning: Bob adds his coin to the side where the straight plate somewhere on his left (doesn't have to be the neighbouring plate, just the first plate on his side that doesn't have exactly $1$ coin) or on his right has $0$ coins to prevent forming a $21112$. or something like this This will always prevent two $2$'s to be coupled (it will always add a plate with $0$ coins between them). Thus, no couples can be formed, so we don't even bother looking at the situation where the straight plates on Bob's left hand side and right hand side are both plates with $2$ coins, as this situation is impossible with our previous method (and because we start with $1$ coin in each plate).