Ben starts with an integer greater than $9$ and subtracts the sum of its digits from it to get a new integer. He repeats this process with each new integer he gets until he gets a positive $1$-digit integer. Find all possible $1$-digit integers Ben can end with from this process.
1. Let's denote the initial integer by \( n \). Since \( n \) is greater than 9, we can write \( n \) in terms of its digits. Suppose \( n \) has digits \( d_1, d_2, \ldots, d_k \), then:
   \[
   n = 10^{k-1}d_1 + 10^{k-2}d_2 + \cdots + 10^0d_k
   \]
2. The sum of the digits of \( n \) is:
   \[
   S(n) = d_1 + d_2 + \cdots + d_k
   \]
3. Ben subtracts the sum of the digits from \( n \) to get a new integer:
   \[
   n' = n - S(n)
   \]
4. We need to show that \( n' \) and \( n \) have the same remainder when divided by 9. By the properties of congruences, we know:
   \[
   n \equiv S(n) \pmod{9}
   \]
   This is because each power of 10 is congruent to 1 modulo 9:
   \[
   10^i \equiv 1 \pmod{9} \quad \text{for any integer } i
   \]
   Therefore:
   \[
   n \equiv d_1 + d_2 + \cdots + d_k \equiv S(n) \pmod{9}
   \]
5. Subtracting \( S(n) \) from \( n \):
   \[
   n' = n - S(n) \implies n' \equiv n - S(n) \equiv 0 \pmod{9}
   \]
   This means \( n' \) is a multiple of 9.
6. Ben repeats this process until he gets a positive 1-digit integer. Since \( n' \) is a multiple of 9 and a 1-digit integer, the only possible 1-digit multiple of 9 is 9 itself.

Conclusion:
\[
\boxed{9}
\]