We consider the degree of polynomial $P$:

Case 1: If $\deg P = 0$ then $P(x) \equiv c$ for some $c \in \mathbb{Z}$, the given condition becomes $S(c) = c$ which holds if and only if $1 \leq c \leq 9$.

Case 2: If $\deg P = 1$. We notice that $S(m+n) \leq S(m) + S(n)$ for all positive integers $m, n$ and the equality occurs when there is no carry in the addition $m+n$. Let $P(x) = a x + b$ for $a, b \in \mathbb{Z}$ and $a \neq 0$. Since $S(P(n)) = P(S(n))$ for $n \geq 2017$ then $a$ must be positive. The given condition becomes $S(an + b) = a S(n) + b$ for $n \geq 2017$. Putting $n = 2025$ and $n = 2020$, we get
$$S(2025 a + b) - S(2020 a + b) = (a S(2025) + b) - (a S(2020) + b) = 5a.$$
We also have
$$S(2025 a + b) = S((2020 a + b) + 5a) \leq S(2020 a + b) + S(5a).$$
These imply that $S(5a) \geq 5a$, as $a \geq 1$, this holds only when $a = 1$. Then we have $S(n + 1 + b) - S(n + b) = S(n + 1) - S(n)$.

1. If $b > 0$ then choose $n$ such that $n + 1 + b = 10^k$ for some $k$ big enough then all digits of $n + b$ are $9$ then the left hand side is $1 - 9k$. Also note that $n$ is a positive integer less than $10^k - 1$, then $S(n) < 9k$, which means $S(n + 1) - S(n) \geq 1 - (9k - 1) = 2 - 9k > 1 - 9k$, a contradiction.

2. If $b < 0$, similarly, we also come to a contradiction.

Thus $b = 0$ and $P(x) = x$, which is trivially satisfied.

Case 3: If $\deg P = d \geq 2$ then the leading term of $P$ is $a_d n^d$ with $a_d \neq 0$, then similarly we have $a_d > 0$. Choose $n = 10^k - 1$ for some $k \in \mathbb{Z}^+$ big enough, then $S(P(n)) = P(9k)$.

Since $P(9k)$ grows approximately as $(9k)^d$ while $S(P(n))$ grows approximately as a constant multiple of $k$, the given equality cannot hold for sufficiently large $k$ since $d \geq 2$.

In conclusion, $P(x) = c$ for some $c \in \{1,2,3, \ldots, 9\}$ or $P(x) = x$, $\forall x$. $\square$