$M = i (i \geq 2N-1)$ Let group A is set of players with player number $2$ to $N+1$ If player 1 is assigned court $N$ and 2 player of group A is assigned court $2$ and 1 player of group A is assigned at court $k$ ($1 \leq k \leq N-1, k \ne 2$) at initial pairing ( $2 \leq i, j \leq N+1$) then after $N-1$ round end player 1 is at court 1 and there are 2 player of group A is at court 2 and 1 player of group A is at court $k$ ($3 \leq k \leq N$) and after $N-1+L$ round end $L \leq N-2$ player 1 is at court 1 and there are 2 player of group A is at court 2 and 1 player of group A is at court $k$ ($1 \leq k \leq N$, $k \ne 2$, $k \ne N-L+1$) and after $2N-2$ round end player 1 is at court 1 and there are 1 player of group A is at court $k$ ($1 \leq k \leq N$) So $M \ne i (i \leq 2N-2)$ After $N-1$th round player 1 is assigned court 1 and doesn't move anymore. After $L$th round $L \geq N-1$ If there are court $k$ with one gourp A player and court $k+1$ with no group A player then court $k$ has no group A player next round. And if there are court $k$ with two group A player and court $k+1$ with no group A player then court $k$ has one group A player next round. So After $2N-2$th round there are no court with no group A player This means all group A player changes court after the $M$th round ($M \geq 2N-1$)

Sorry for the revive, but I'm confused about a part of the above solution. Quote: If player 1 is assigned court $N$ and 2 player of group A is assigned court $2$ and 1 player of group A is assigned at court $k$ ($1 \leq k \leq N-1, k \ne 2$) at initial pairing ( $2 \leq i, j \leq N+1$) then after $N-1$ round end player 1 is at court 1 and there are 2 player of group A is at court 2 and 1 player of group A is at court $k$ ($3 \leq k \leq N$) If the initial pairing looks like $$\begin{bmatrix}A & A & A & \cdots & A & 1 \\ B & A & B & \cdots & B & B \end{bmatrix},$$after $N-1$ moves, the pairing becomes $$\begin{bmatrix}1 & A & A & \cdots & A & A \\ A & B & B & \cdots & B & B\end{bmatrix},$$unless I'm misinterpreting the problem. In this case, it is evident that all of the $A$s move in the $N$th round.

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and the correct answer is $M \geq N+1$ for all $N \geq 3$. (The $N = 2$ has the pathological answer of $2$ but as $N = 2$ can be bashed by hand, assume $N \geq 3$ thorough the solution). Call a configuration of the players tasty at some point of time if after the current match the players $2, 3, \cdots, N+1$ switch places. The construction that is achieved by this: Put persons ranked $2N, 2N-1$ in the first court. And for $2 \leq i \leq N$th court, put the persons $i-1, N+i-2$ on that. This becomes tasty only from $N+1$th match (while this is not immediately obvious why this is true, it would be obvious after reading the proof of the bound's optimality). Call the player ranked $1$ as Federer. Note that if Federer is at the court number $j$ with $j > 1$ at match $t$, after match $t$ he will be at court number $j-1$. So if he was at court number $j$ at the beginning of the tournament, he will be at court number $1$ after exactly $j-1$ matches. Also note that if he ever steps into the court numbered $1$, he will never switch courts ever. So regardless of the initial distribution, Federer comes at court number $1$ and stops switching after $N-1$th match. So we need not worry about him. Now buy $N+1$ blue masks (which cover the faces) and $N-1$ red masks (which cover the faces). At the very beginning of the tournament, put a blue mask on the players $\{1, \cdots, N+1 \}$ and a red mask on $\{N+2, \cdots, 2N \}$. Note that if after match $t$ Federer is at first court In every $i \in \{2, \cdots, N \}$th court, there's exactly one red masked person. , then the configuration is tasty after match $t$ and will forever be after all $M \geq t$. Since after $N-1$th match, the condition (1) is always satisfied, we need to make sure the condition (2) is always satisfied after $N+1$th match, which is the main difficulty of the problem. Also the following part of the proof is most probably not comprehensible/wrong, but I have tried to make it understandable by adding an example of what I mean. Assume the persons are indistinguishable - i.e only thing that distinguishes person A from person B is the color of their masks, and two persons with same color of masks are not distinguishable from each other. Let $a_{i, t}$ denote the number of redmasked people at court $i$ after the $t$-th match. Now go to a big ground, and mark a big circle and $N$ points on it, $p_1, \cdots, p_n$ in that order. Dig one hole each at $p_2, \cdots, p_n$ with dimensions $1 \times 1 \times 1 \text{ metre}$ (there's no hole at $p_1$). Also by $N-1$ cubical red boxes with dimensions $1 \times 1 \times 1 \text{ metre}$. At the beginning of the tournament, put $a_{i, 0}$ boxes on $p_i$ (they will fall into the hole, and maybe stack up on each other). After each match, do the following: for $ 1 \leq i \leq N$, if there's a box at $p_{i}$ (indices modulo $N$) which is at ground level (those which fall inside the hole are never picked up, and if there's two boxes at $p_{i}$ above the ground level (which is only possible if $p_{i} = p_0$) then move only the bottom one), move it to $p_{i-1}$ (indices modulo $N$). Let $b_{i, t}$ denote the total number of boxes at $p_i$ after match $t$. It can be easily verified that $b_{i,t} = a_{i,t}$. So we need to prove that for all $t \geq N+1, 2 \leq i \leq N$, $b_{i, t}= 1$. In other words, we need to prove that any box falls into a hole within $N$ matches. This is not hard. Note that a box can't go round through all the entire points and come back to the original position without falling into a hole since that would imply the existence of atleast $N$ red boxes. So starting from $p_i$, it can atmax go to $p_{i+1}$ (going like this $p_{i} \rightarrow p_{i-1} \rightarrow p_{i-2} \cdots p_{i+1}$ and then finally falling down at $p_{i+1}$, so total after $N-1$ matches). Also till the box fallls down in a hoe, the box moves one step forward. The only exception is that it might have to "pause" for a round at $p_1$ (since there might be two boxes and it might be the upper one). So total any box takes at maximum $N-1 + 1 = N$ matches before permanently falling down to a hole, which is what we wanted to prove. Here's an example of the above process (Note $|AB|CD|EF|GH|$ means at that round $A,B$ are playing at court $1$, $C,D$ playing at court $2$ etc. Also $R$ means red mask, $B$ means blue mask: $|78|14|25|36| \overset{\text{mask}}{\rightarrow} |RR|BB|BB|BR|$. That's why note in (1), there's two red boxes at $p_1$, none in $p_2, p_3$ and one in $p_4$. Also note that $A$ doesn't move since there's $B$ below $A$ but above the ground level and that's why only $B$ is moved $|17|24|35|68| \overset{\text{mask}}{\rightarrow} |BR|BB|BB|RR|$ . Both $A$ and $B$ move. Now $C$ doesn't move since it has fallen into a hole. $|12|43|56|87| \overset{\text{mask}}{\rightarrow} |BB|BB|BR|RR|$. $|13|45|67|82| \overset{\text{mask}}{\rightarrow} |BB|BB|RR|RB|$ $|14|56|27|38| \overset{\text{mask}}{\rightarrow} |BB|BR|BR|BR|$. This is now tasty.

An easy way to see this is that 1 cannot lose and we see that for all 2 to n+1 to move we want them to win and only lose against 1. So as all should move all have to be in separate courts. So the final config will have 1 in court 1 and n+2 ti 2n in other courts with 2 to n+1 rotating. So we see that the only way we can assure 1 is not moving is by n-1 rounds and so min is n-1. Now see the fact that 2n-1 can only move if it fights 2n and 2n-2 can only move if it fights 2n-1 and 2n . ( i guess you can see where this is heading). We see that after inital 2 rounds we can assure that 2n-1 and 2n are stable or have a fixed court. ( see and do the reason why it has to be 2 not 1). And continuing this we get that after N-1 rounds we can assure ourselves that all will be stable. So after Nth round all will move so M=>N .