Find all the real numbers $ \alpha$ satisfy the following property: for any positive integer $ n$ there exists an integer $ m$ such that $ \left |\alpha\minus{}\frac{m}{n}\right|<\frac{1}{3n}$.
To solve the problem, we need to find all real numbers \(\alpha\) that satisfy the given property: for any positive integer \(n\), there exists an integer \(m\) such that \(\left|\alpha - \frac{m}{n}\right| < \frac{1}{3n}\).

1. **Case 1: \(\alpha\) is an integer**

   Let \(\alpha = k\) where \(k\) is an integer. For any positive integer \(n\), we can choose \(m = kn\). Then,
   \[
   \left| \alpha - \frac{m}{n} \right| = \left| k - \frac{kn}{n} \right| = \left| k - k \right| = 0
   \]
   Since \(0 < \frac{1}{3n}\) for any positive integer \(n\), the condition is satisfied. Therefore, any integer \(\alpha\) satisfies the property.

2. **Case 2: \(\alpha\) is not an integer**

   Suppose \(\alpha\) is not an integer. We need to show that there exists a positive integer \(n\) such that for any integer \(m\), \(\left|\alpha - \frac{m}{n}\right| \geq \frac{1}{3n}\).

   Since \(\alpha\) is not an integer, it can be written as \(\alpha = k + \beta\) where \(k\) is an integer and \(0 < \beta < 1\). For any positive integer \(n\), consider the fractional part of \(n\alpha\), denoted by \(\{n\alpha\}\). The fractional part \(\{n\alpha\}\) is given by:
   \[
   \{n\alpha\} = n\alpha - \lfloor n\alpha \rfloor
   \]
   where \(\lfloor x \rfloor\) denotes the floor function of \(x\).

   Since \(\alpha\) is irrational, \(\{n\alpha\}\) is uniformly distributed in \((0, 1)\). Therefore, for any positive integer \(n\), there exists an integer \(m\) such that:
   \[
   \left| \alpha - \frac{m}{n} \right| = \left| \alpha - \frac{\lfloor n\alpha \rfloor}{n} \right| = \left| \frac{n\alpha - \lfloor n\alpha \rfloor}{n} \right| = \frac{\{n\alpha\}}{n}
   \]
   Since \(\{n\alpha\}\) is uniformly distributed in \((0, 1)\), there will be values of \(n\) for which \(\{n\alpha\}\) is arbitrarily close to 0 or 1. However, for any \(n\), \(\{n\alpha\}\) cannot be exactly 0 or 1 because \(\alpha\) is irrational. Therefore, \(\frac{\{n\alpha\}}{n}\) can be made arbitrarily small but never zero.

   Hence, for any positive integer \(n\), there exists an integer \(m\) such that:
   \[
   \left| \alpha - \frac{m}{n} \right| < \frac{1}{3n}
   \]
   This implies that the property holds for any real number \(\alpha\), whether it is an integer or not.

Conclusion:
The property is satisfied for all real numbers \(\alpha\).

The final answer is \(\boxed{\text{all real numbers}}\)