Four. (50 points) Let $A=\{0,1, \cdots, 2016\}$. If a surjective function $f: \mathbf{N} \rightarrow A$ satisfies: for any $i \in \mathbf{N}$,
$$
f(i+2017)=f(i),
$$

then $f$ is called a "harmonious function".
$$
\begin{array}{l}
\text { Let } f^{(1)}(x)=f(x), \\
f^{(k+1)}(x)=f\left(f^{(k)}(x)\right)\left(k \in \mathbf{N}_{+}\right) .
\end{array}
$$

Suppose the "harmonious function" $f$ satisfies the condition: there exists a positive integer $M$, such that
(1) When $m<M$, if $i, j \in \mathbf{N}$, $i \equiv j+1(\bmod 2017)$, then $f^{(m)}(i)-f^{(m)}(j) \not \equiv \pm 1(\bmod 2017)$; (2) If $i, j \in \mathbf{N}, i \equiv j+1(\bmod 2017)$, then $f^{(M)}(i)-f^{(M)}(j) \equiv \pm 1(\bmod 2017)$. Find the maximum possible value of $M$.
On the one hand, note that 2017 is a prime number.
Let $g$ be a primitive root modulo 2017, then the half-order of $g$ modulo 2017 is 1008.
Let $f(i) \equiv g(i-1)+1(\bmod 2017)$.
Since $(g, 2017)=1$, $g(i-1)+1$ traverses a complete residue system modulo 2017.
Thus, the mapping $f: \mathbf{N} \rightarrow A$ is surjective.
Also, $f(i+2017) \equiv f(i)(\bmod 2017)$, i.e.,
$$
f(i+2017)=f(i) \text {, }
$$

Therefore, the defined $f$ is a "harmonious mapping".
According to the definition of $f$,
$$
f^{(k)}(i) \equiv g^{k}(i-1)+1(\bmod 2017) .
$$

At this point, from $i \equiv j+1(\bmod 2017)$, we get
$$
\begin{array}{l}
\left(g^{M}(i-1)+1\right)-\left(g^{M}(j-1)+1\right) \\
\equiv \pm 1(\bmod 2017) \\
\Leftrightarrow g^{M}(i-j) \equiv \pm 1(\bmod 2017) \\
\Leftrightarrow g^{M} \equiv \pm 1(\bmod 2017) .
\end{array}
$$

Noting that the half-order of $g$ modulo 2017 is 1008.
Thus, $1008 \mid M$.
Therefore, the required $M_{\max } \geqslant 1008$.
On the other hand, construct a convex 2017-gon $A_{0} A_{1} \cdots A_{2016}$, denoted as graph $G$. Connect lines according to the following rule: if $1 \leqslant m<M$,
$$
i \in A, f^{m}(i)=a, f^{m}(i+1)=b \text {, then connect segment } A_{a} A_{b} \text {. }
$$

Clearly, the connected segments are diagonals of graph $G$, and the connected segments are not repeated. Otherwise, if there exist $i, j \in A$, and $1 \leqslant q<p<M$, such that
$$
\begin{array}{l}
f^{(p)}(i)=f^{(q)}(j), \text { and } \\
f^{(p)}(i+1)=f^{(q)}(j+1) \\
\text { or } f^{(p)}(i+1)=f^{(q)}(j-1) . \\
\quad \text { then }\left\{\begin{array}{l}
f^{(p-q)}(i)=j, \\
f^{(p-q)}(i+1) \equiv j+1 \text { or } j-1(\bmod 2017) .
\end{array}\right.
\end{array}
$$

Noting that $0<p-q<M$, this contradicts the given conditions. Hence, the connected diagonals are not repeated.

Since a total of $2017(M-1)$ segments are connected, and the convex 2017-gon $G$ has $2017 \times 2017$ segments, thus,
$2017(M-1) \leqslant 2017 \times 1007$
$\Rightarrow M \leqslant 1008$.
In summary, the maximum value of $M$ is 1008.
(Zhu Xiaoguang, Tian Kaibin, Pan Chenghua, Wenwu Guanghua Mathematics Studio, 215128)