For this editorial, denote $f(x,y)$ as the interactor's response after making a query with $l=x$ and $r=y$. Also, $a_1,a_2,\dots ,a_N$ is the sequence which we are trying to find. Finally, let $\oplus$ denote the bitwise XOR operator.

$M=1$

If we do not restrict ourselves to maximising the value of $M$, then there are many possible solutions, one of which will be described here.

First, we can query $f(1,2)$, $f(1,3)$ and $f(2,3)$, which allows us to find $a_1$, $a_2$ and $a_3$.

Then, for each $x$ from $4$ up to $N$, we can find $a_x$ by calculating $a_{x-1}\oplus f(x-1,x)$.

$M=\lfloor \frac{N-1}{2}\rfloor$

Again, there are many constructions which produce a value of $M=\lfloor \frac{N-1}{2}\rfloor$. It turns out that this is the maximum possible value of $M$. A proof of this will follow after an example of a possible construction.

First, find the value of $f(i,N)$ for $1\le N\le \lceil \frac{N+1}{2}\rceil$. This lets us work out the value of $a_x$ for $1\le x\le \lceil \frac{N-1}{2}\rceil$, since $a_x=f(x,N)\oplus f(x+1,N)$.

Then, we can work out the value of $a_x$ for $\lceil \frac{N+1}{2}\rceil \le x<N$ with one additional query for each $x$, since $a_x=f(1,x)\oplus a_1\oplus a_2\oplus \cdots \oplus a_{x-1}$. Cumulative prefix XOR sums can be used to speed up this computation.

Finally, we can work out $a_N$ as equal to $f(1,N)\oplus a_1\oplus a_2\oplus \cdots \oplus a_{N-1}$. Note that $f(1,N)$ has already been found from the first step, so this does not require an extra query.

This solves the problem with $N$ queries and $M=\lfloor \frac{N-1}{2}\rfloor$.

Now, let's consider why this value of $M$ is maximal. Let $p_x$ represent the prefix XOR sequence, where $p_x=a_1\oplus a_2\oplus \cdots \oplus a_x$. For convenience, let $p_0=0$. Note that being able to work out sequence $p$ is a necessary and sufficient condition for being able to work out sequence $a$, so we will turn our attention to finding sequence $p$.

We can see that $f(l,r)=p_{l-1}\oplus p_r$. Therefore, if we know $f(l,r)$ and one of $p_{l-1}$ or $p_r$, then we can work out the other value out of $p_{l-1}$ or $p_r$.

This motivates the following idea: create a graph containing $N+1$ nodes labelled $0,1,2,\dots ,N$. Whenever we make a query asking for $f(l,r)$, then add a bidirectional edge between $l-1$ and $r$. Then, two nodes $x$ and $y$ are connected if and only if knowing the value of $p_x$ allows us to figure out $p_y$, and vice versa.

After performing $N$ queries, we require the graph to be connected; otherwise, there is no unique solution. Since there are $N+1$ nodes and $N$ edges, the edges must form a spanning tree of this graph. This also explains why the minimal number of queries to find sequence $a$ is $N$.

Recall that the value of $M$ is the minimum value of $|x-y|-1$, for all edges $(x,y)$ in the graph. Now, consider node $\lfloor \frac{N}{2}\rfloor$. Since the edges form a spanning tree, this node must be connected to another node. However, the minimum value of $|\lfloor \frac{N}{2}\rfloor -x|-1$ for $0\le x\le N$ is $\lfloor \frac{N-1}{2}\rfloor$. This proves that $M$ cannot exceed $\lfloor \frac{N-1}{2}\rfloor$. On the other hand, we have already shown via construction that this value of $M$ is attainable, so $\lfloor \frac{N-1}{2}\rfloor$ is the maximum possible value of $M$.

This graph idea also lends itself to a different construction. First, generate any spanning tree of the $N+1$ nodes such that $|x-y|-1\ge \lfloor \frac{N-1}{2}\rfloor$ for all edges $(x,y)$. For each edge $(x,y)$ in this graph with $x<y$, let it have a weight of $f(x+1,y)$. We know that $p_0=0$. Performing a DFS/BFS on this graph, we can work out the entire sequence $p$, as $p_x$ is simply the XOR sum of all edges on the path from node $0$ to node $x$. Once we know $p$, we can easily find sequence $a$, since $a_x=p_{x-1}\oplus p_x$ for all $1\le x\le N$.