4. If a rational number has a finite number of digits in its decimal representation, it is called "short". For positive integers $m, t$, if there exists an integer $c \in\{1,2, \cdots, 2017\}$ such that $\frac{10^{t}-1}{c m}$ is short, and for any $k \in\{1, 2, \cdots, t-1\}, \frac{10^{k}-1}{c m}$ is not short, then $t$ is called "m wonderful". Let $S(m)$ be the set of m wonderful numbers. For $m=1,2, \cdots$, find the maximum number of elements in $S(m)$.
4. 807.

First, $x \in \mathbf{Q}$ is short if and only if there exist non-negative integers $a, b$ such that $2^{a} 5^{b} x \in \mathbf{Z}$.

In fact, if $x$ is short, then there exist integers $n$ and non-negative integers $k$ such that $x=\frac{n}{10^{k}}$. Taking $a=b=k$ suffices; if there exist non-negative integers $a, b$ such that $2^{a} 5^{b} x = q \in \mathbf{Z}$, then $x=\frac{2^{b} 5^{a} q}{10^{a+b}}$ is short.

If $m=2^{a} 5^{b} s\left(s \in \mathbf{Z}_{+}\right.$ and $\left.(s, 10)=1\right)$, then $\frac{10^{t}-1}{m}$ is short if and only if $s \mid\left(10^{t}-1\right)$.
Without loss of generality, assume $(m, 10)=1$.
Define
$C=\left\{c \in \mathbf{Z}_{+} \mid 1 \leqslant c \leqslant 2017, (c, 10)=1\right\}$.
Then $m$ is a wonderful number if it is the smallest positive integer $t$ such that for some integer $c \in C$, we have
$$
10^{t} \equiv 1(\bmod c m),
$$

i.e., the set of orders of 10 modulo $\mathrm{cm}$.
This indicates that,
$$
S(m)=\left\{\operatorname{ord}_{c m} 10 \mid c \in C\right\},
$$

where $\operatorname{ord}_{i} j$ denotes the order of the positive integer $j$ modulo the positive integer $i$.
Since there are $4 \times 201+3=807$ integers $c (1 \leqslant c \leqslant 2017)$ satisfying $(c, 10)=1$, i.e.,
$$
c \equiv 1,3,7,9(\bmod 10),
$$

we have $|S(m)| \leqslant |C|=807$.
We now prove: there exists a positive integer $m$ such that
$|S(m)|=807$.
Let $P=\{p$ be a prime $11<p \leqslant 2017$,
$$
p \notin\{2,5\}\} \text {. }
$$

Choose a positive integer $\alpha$ such that for each $p \in P$, we have $p \mid\left(10^{\alpha}-1\right)$ (e.g., take $\alpha=\varphi(T)$, where $T$ is the product of all primes in $P$). Set $m=10^{\alpha}-1$.
First, we prove a lemma.
Lemma For each $c \in C$, we have
$$
\operatorname{ord}_{c m} 10=c \alpha \text {. }
$$

Proof Clearly, $\operatorname{ord}_{m} 10=\alpha$.
Let $t=\operatorname{ord}_{c m} 10$. Then
$$
c m\left|\left(10^{t}-1\right) \Rightarrow m\right|\left(10^{t}-1\right) \Rightarrow \alpha \mid t \text {. }
$$

Thus, there exists a positive integer $k$ such that $t=k \alpha$.
Let $v_{p}(n)$ denote the power of the prime $p$ in the prime factorization of the positive integer $n$.
For a positive integer $l$ and $p \in P$, by
$$
10^{\alpha} \equiv 1(\bmod m),
$$

we have $v_{p}\left(10^{l \alpha}-1\right)=v_{p}\left(\left(10^{\alpha}\right)^{t}-1\right)$
$$
\begin{array}{l}
=v_{p}\left(10^{\alpha}-1\right)+v_{p}(l) \\
=v_{p}(m)+v_{p}(l) .
\end{array}
$$

Then $c m \mid\left(10^{k \alpha}-1\right)$
$\Leftrightarrow$ for all $p \in P$, we have
$$
v_{p}(\mathrm{~cm}) \leqslant v_{p}\left(10^{k \alpha}-1\right)
$$
$\Leftrightarrow$ for all $p \in P$, we have
$$
v_{p}(m)+v_{p}(c) \leqslant v_{p}(m)+v_{p}(k)
$$
$\Leftrightarrow$ for all $p \in P$, we have
$$
v_{p}(c) \leqslant v_{p}(k)
$$
$\Leftrightarrow c \mid k$.
Thus, the $k$ satisfying this condition must be $k=c$.
Therefore, $\operatorname{ord}_{c m} 10=c \alpha$.
The lemma is proved.
By the lemma, we know $|S(m)|=|C|=807$.