We will show that the desired maximum value for $m$ is $n(n-1)$. First, let us show that $m \leq n(n-1)$ always holds for any sequence $P_{0}, P_{1}, \cdots, P_{m+1}$ satisfying the conditions of the problem. Call a point a turning point if it coincides with $P_{i}$ for some $i$ with $1 \leq i \leq m$. Let us say also that 2 points $\{P, Q\}$ are adjacent if $\{P, Q\}=\left\{P_{i-1}, P_{i}\right\}$ for some $i$ with $1 \leq i \leq m$, and vertically adjacent if, in addition, $P Q$ is parallel to the $y$-axis. Any turning point is vertically adjacent to exactly one other turning point. Therefore, the set of all turning points is partitioned into a set of pairs of points using the relation of "vertical adjacency". Thus we can conclude that if we fix $k \in\{1,2, \cdots, n\}$, the number of turning points having the $x$-coordinate $k$ must be even, and hence it is less than or equal to $n-1$. Therefore, altogether there are less than or equal to $n(n-1)$ turning points, and this shows that $m \leq n(n-1)$ must be satisfied. It remains now to show that for any positive odd number $n$ one can choose a sequence for which $m=n(n-1)$. We will show this by using the mathematical induction on $n$. For $n=1$, this is clear. For $n=3$, choose $P_{0}=(0,1), \quad P_{1}=(1,1), \quad P_{2}=(1,2), \quad P_{3}=(2,2), \quad P_{4}=(2,1), \quad P_{5}=(3,1), \quad P_{6}=(3,3), \quad P_{7}=(4,3)$. It is easy to see that these points satisfy the requirements. Let $n$ be an odd integer $\geq 5$, and suppose there exists a sequence satisfying the desired conditions for $n-4$. Then, it is possible to construct a sequence which gives a configuration indicated in the following diagram, where the configuration inside of the dotted square is given by the induction hypothesis: By the induction hypothesis, there are exactly $(n-4)(n-5)$ turning points for the configuration inside of the dotted square, and all of the lattice points lying outside of the dotted square except for the 4 points $(n, 2),(n-1, n-2),(2,3),(1, n-1)$ are turning points. Therefore, the total number of turning points in this configuration is $(n-4)(n-5)+\left(n^{2}-(n-4)^{2}-4\right)=n(n-1)$ showing that for this $n$ there exists a sequence satisfying the desired properties, and thus completing the induction process.