For any positive integer $k, S(k)$ is the sum of the digits of $k$ in its decimal representation. Find all integer-coefficient polynomials $P(x)$ such that for any positive integer $n \geqslant 2$016, $P(n)$ is a positive integer, and
$$
S(P(n))=P(S(n)) \text {. }
$$
1. $P(x)=c$ (integer $c$ satisfies $1 \leqslant c \leqslant 9$) or $P(x)=x$.
Below, we discuss the degree of $P(x)$ in different cases.
(1) If $P(x)$ is a constant polynomial.
Let $P(x)=c$ ($c$ is an integer and a constant). Then equation (1) becomes $S(c)=c$.
Thus, the equation holds if and only if $1 \leqslant c \leqslant 9$.
(2) If $\operatorname{deg} P=1$.
First, for any positive integers $m, n$, we have
$$
S(m+n) \leqslant S(m)+S(n),
$$
with equality holding if and only if $m+n$ does not carry over.
Let $P(x)=a x+b$ ($a, b$ are integers, $a \neq 0$).
For sufficiently large $n$, $P(n)$ is a positive integer, so $a$ is a positive integer. Thus, for any positive integer $n \geqslant 2016$, equation (1) becomes
$$
\begin{array}{l}
S(a n+b)=a S(n)+b. \\
\text{Let } n=2025, 2020, \text{ we get} \\
S(2025 a+b)-S(2020 a+b) \\
=(a S(2025)+b)-(a S(2020)+b) \\
=5 a.
\end{array}
$$

Second, by equation (2), we know
$$
\begin{array}{l}
S(2025 a+b)=S((2020 a+b)+5 a) \\
\leqslant S(2020 a+b)+S(5 a).
\end{array}
$$

This indicates that $5 a \leqslant S(5 a)$.
Since $a$ is a positive integer, the inequality holds only when $a=1$. Thus, for any positive integer $n \geqslant 2016$, equation (1) becomes
$$
\begin{array}{l}
S(n+b)=S(n)+b. \\
\text{Hence, } S(n+1+b)-S(n+b) \\
=(S(n+1)+b)-(S(n)+b) \\
=S(n+1)-S(n).
\end{array}
$$

If $b>0$, then there exists a positive integer $n$ such that for sufficiently large positive integer $k$, we have $n+1+b=10^{k}$.
Notice that all digits of $n+b$ are 9.
Then the left side of equation (3) equals $1-9 k$.
By $n0$.
In equation (1), let $n=10^{k}-1$. Then
$$
S(P(n))=P(9 k).
$$

Notice that $P(n)$ increases at the order of $n^{d}$.
Thus, $S(P(n))$ increases at the order of a constant times $k$.

Also, $P(9 k)$ increases at the order of $k^{d}$, and since $d \geqslant 2$, equation (4) cannot hold when $k$ is sufficiently large.
In summary, $P(x)=c(1 \leqslant c \leqslant 9)$ or $P(x)=x$.