Let $n$ be a positive four-digit integer, and let $m$ be the eight-digit integer formed by repeating $n$.
Then $m = 10000n + n = 10001n$.
We can see that every eight-digit integer formed by repeating a positive four-digit integer is a multiple of $10001$.
Therefore, the highest common factor of all these integers is $\boxed{10001}$.
The answer is: 10001
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
What is the highest common factor of all eight-digit integers that are formed by repeating a positive four-digit integer?

### Answer:
