A rational number $ x$  is called [i]good[/i] if it satisfies: $ x\equal{}\frac{p}{q}>1$ with $ p$, $ q$ being positive integers, $ \gcd (p,q)\equal{}1$ and there exists constant numbers $ \alpha$, $ N$ such that for any integer $ n\geq N$, \[ |\{x^n\}\minus{}\alpha|\leq\dfrac{1}{2(p\plus{}q)}\] Find all the good numbers.
To find all the good numbers \( x \) that satisfy the given conditions, we will follow a detailed step-by-step approach.

1. **Definition and Initial Setup:**
   A rational number \( x = \frac{p}{q} > 1 \) is called *good* if there exist constants \( \alpha \) and \( N \) such that for any integer \( n \geq N \),
   \[
   |\{x^n\} - \alpha| \leq \frac{1}{2(p+q)}
   \]
   where \( \{x^n\} \) denotes the fractional part of \( x^n \).

2. **Expression for \( x^n \):**
   For \( n \geq N \), we can write:
   \[
   x^n = a_n + \alpha + \beta_n
   \]
   where \( a_n \in \mathbb{Z} \) and \( |\beta_n| \leq \frac{1}{2(p+q)} \).

3. **Bounding \( \beta_n \):**
   The equality \( |\beta_n| = \frac{1}{2(p+q)} \) can only occur a finite number of times if \( q > 1 \). For sufficiently large \( n \), we have:
   \[
   |\beta_n| < \frac{1}{2(p+q)}
   \]

4. **Integer Condition:**
   Consider the expression:
   \[
   qa_n - pa_{n-1} = \alpha (p - q) + \beta_{n-1} p - \beta_n q
   \]
   Since \( \alpha (p - q) + \beta_{n-1} p - \beta_n q \) is an integer, for large \( n \), we have:
   \[
   |\beta_{n-1} p - \beta_n q| < \frac{1}{2}
   \]
   This implies that \( \alpha (p - q) + \beta_{n-1} p - \beta_n q \) must be the nearest integer to \( \alpha (p - q) \), which is unique since it differs from \( \alpha (p - q) \) by less than \( \frac{1}{2} \).

5. **Constant Integer \( c \):**
   Let this integer be \( c \). Thus, for large \( n \), we have:
   \[
   qa_n = pa_{n-1} + c
   \]

6. **Sequence Analysis:**
   Define \( b_n = (p - q)a_n + c \). The sequence \( \{b_i\} \) (for \( i \geq n_0 \)) is a sequence of positive integers such that:
   \[
   qb_n = pb_{n-1}
   \]

7. **Divisibility Argument:**
   For a given \( n \), \( q^k \) divides \( b_n \) for all \( k \) because:
   \[
   q^k b_{n+k} = p^k b_n
   \]
   and \( \gcd(p, q) = 1 \). This implies that \( q = 1 \).

8. **Conclusion:**
   If \( q = 1 \), then \( x = p \) is an integer. Therefore, \( x \) must be an integer greater than 1. Conversely, if \( x \) is an integer greater than 1, it satisfies the given condition, making it a good number.

The final answer is \( \boxed{ x } \) is an integer greater than 1.