Positive integers are colored either blue or red such that if $a,b$ have the same color and $a-10b$ is a positive integer then $a-10b, a$ have the same color as well. How many such coloring exist?
1. **Understanding the problem**: We need to find the number of ways to color positive integers either blue or red such that if two integers \(a\) and \(b\) have the same color and \(a - 10b\) is a positive integer, then \(a - 10b\) and \(a\) must have the same color.

2. **Analyzing the given solution**: The solution suggests that there are infinitely many such colorings. Specifically, for any positive integer \(k\), if we color \(k\) and \(11k\) blue and all other integers red, the condition is satisfied.

3. **Verification of the given solution**:
    - Let \(k\) be a positive integer.
    - Color \(k\) and \(11k\) blue.
    - Color all other integers red.
    - Check the condition: If \(a\) and \(b\) have the same color and \(a - 10b\) is a positive integer, then \(a - 10b\) and \(a\) must have the same color.
        - If \(a = k\) and \(b = k\), then \(a - 10b = k - 10k = -9k\), which is not positive.
        - If \(a = 11k\) and \(b = k\), then \(a - 10b = 11k - 10k = k\), which is positive. Since \(k\) and \(11k\) are both blue, the condition is satisfied.
        - For any other \(a\) and \(b\) that are red, \(a - 10b\) will also be red, satisfying the condition.

4. **Generalization**: The given solution can be generalized to any set of integers where the condition holds. For example, if we choose any integer \(m\) and color \(m\) and \(11m\) blue, and all other integers red, the condition will still be satisfied.

5. **Conclusion**: Since we can choose any positive integer \(k\) and color \(k\) and \(11k\) blue while coloring all other integers red, there are infinitely many such colorings.

The final answer is \( \boxed{ \text{infinitely many such colorings} } \).