Let $ \alpha,\beta$ be real numbers satisfying $ 1 < \alpha < \beta.$ Find the greatest positive integer $ r$ having the following property: each of positive integers is colored by one of $ r$ colors arbitrarily, there always exist two integers $ x,y$ having the same color such that $ \alpha\le \frac {x}{y}\le\beta.$
1. **Claim**: The greatest positive integer \( r \) having the desired property is \( r = 2 \).

2. **Proof for \( r = 2 \)**:
    - Consider the interval \((\alpha, \beta)\) where \(1 < \alpha < \beta\).
    - Let \(\gamma = \beta - \alpha\).
    - By the Archimedean property of the reals, there exists an integer \(N\) such that \(N\gamma > \lceil \alpha \rceil + 3\).
    - This implies that there are at least \(\lceil \alpha \rceil + 1\) integers in the range \((N\alpha, N\beta)\).
    - Suppose for contradiction that for all \(y\), there is no \(x\) such that \(\frac{x}{y} \in (\alpha, \beta)\) and \(x\) and \(y\) are the same color.
    - This means all integers in the range \((N\alpha, N\beta)\) are the opposite color of \(N\).
    - Similarly, all integers in the range \(((N+1)\alpha, (N+1)\beta)\) are the opposite color of \(N+1\).
    - Since \(N(\beta - \alpha)\) is chosen to be larger than \(\lceil \alpha \rceil + 3\), these two intervals will have an integer in their intersection.
    - Thus, \(N\) and \(N+1\) must have the same color, leading to a contradiction.
    - Therefore, \(r = 2\) has the desired property.

3. **Counterexample for \( r = 3 \)**:
    - Consider the coloring pattern: \(123111222333111111111222222222333333333...\), where:
        - \([\frac{3^n - 1}{2}, \frac{3^n - 1}{2} + 3^{n-1})\) is colored 1,
        - \([\frac{3^n - 1}{2} + 3^{n-1}, \frac{3^n - 1}{2} + 2 \cdot 3^{n-1})\) is colored 2,
        - \([\frac{3^n - 1}{2} + 2 \cdot 3^{n-1}, \frac{3^{n+1} - 1}{2})\) is colored 3.
    - The smallest ratio for the same group is:
        \[
        \frac{\frac{3^n - 1}{2} + 3^{n-1} - 1}{\frac{3^n - 1}{2}} = \frac{3^n - 1 + 2 \cdot 3^{n-1} - 2}{3^n - 1} = \frac{\frac{5}{3}(3^n - 1) - \frac{4}{3}}{3^n - 1} < \frac{5}{3}
        \]
    - The smallest ratio for adjacent groups is:
        \[
        \frac{\frac{3^{n+1} - 1}{2} + 2 \cdot 3^n}{\frac{3^{n+1} - 1}{2} - 1} = \frac{3^{n+1} - 1 + 4 \cdot 3^n}{3^{n+1} - 3} = \frac{\frac{7}{3}(3^{n+1} - 3) + 6}{3^{n+1} - 3} > \frac{7}{3}
        \]
    - Thus, choosing \((\alpha, \beta) = (\frac{5}{3}, \frac{7}{3})\) leads to a counterexample for \(r = 3\).

The final answer is \( \boxed{ r = 2 } \)