4. Given that $n$ is a positive integer with $b$ digits, $l, r \in \mathbf{N}$, and $l+r<b$, a positive integer is called a "sub-factor" of $n$ if and only if it divides the positive integer obtained by removing the first $l$ digits and the last $r$ digits of $n$. For example, the sub-factors of 143 are $1, 2, 3, 4, 7, 11, 13, 14, 43$, and 143.

For any positive integer $d$, let $A_{d}$ be the set of positive integers such that $d$ is not a sub-factor of any element in the set. Find all positive integers $d$ such that the set $A_{d}$ is finite.
4. All positive integers coprime with 10.

If an integer is divisible by 2 or 5, then it cannot divide any positive integer of the form $\overline{11 \cdots 1}$. Therefore, for all positive integers $d$ that are not coprime with 10, the set $A_{d}$ is not a finite set.
Let $d$ be a positive integer coprime with 10, and let $b>d$.
We will prove that any positive integer $a_{b} a_{b-1} \cdots a_{1}$ is not in the set $A_{d}$.
Consider the positive integers
$$
a_{1}, \overline{a_{2} a_{1}}, \overline{a_{3} a_{2} a_{1}}, \cdots, \overline{a_{b} a_{b-1} \cdots a_{1}}
$$

under modulo $d$.
Since $b>d$, there exist positive integers $i, j (i>j)$ such that $\qquad$
$\overline{a_{i} a_{i-1} \cdots a_{1}} \equiv \overline{a_{j} a_{j-1} \cdots a_{1}}(\bmod d)$.
Then $\overline{a_{i} a_{i-1} \cdots a_{j+1}} \cdot 10^{j} \equiv 0(\bmod d)$.
Since $d$ is coprime with 10, we have
$d \mid \overline{a_{i} a_{i-1} \cdots a_{j+1}}$.
Thus, $\overline{a_{b} a_{b-1} \cdots a_{1}}$ is not in the set $A_{d}$, which indicates that all positive integers in $A_{d}$ have at most $d$ digits.
Therefore, the set $A_{d}$ is a finite set.