Find all pairs $(a, b)$ of real numbers such that $a \cdot \lfloor b \cdot n\rfloor  = b   \cdot \lfloor  a  \cdot n  \rfloor$ applies to all positive integers$ n$.
(For a real number $x, \lfloor x\rfloor$  denotes the largest integer that is less than or equal to $x$.)
To find all pairs \((a, b)\) of real numbers such that \(a \cdot \lfloor b \cdot n \rfloor = b \cdot \lfloor a \cdot n \rfloor\) for all positive integers \(n\), we will analyze different cases based on the nature of \(a\) and \(b\).

1. **Case 1: \(a\) and \(b\) are not integers**

   Let \(a_1\) and \(b_1\) be the fractional parts of \(a\) and \(b\) respectively. For \(n = 1\), we have:
   \[
   a \cdot \lfloor b \rfloor = b \cdot \lfloor a \rfloor
   \]
   Since \(a\) and \(b\) are not integers, \(\lfloor a \rfloor\) and \(\lfloor b \rfloor\) are integers. Let \(\frac{a}{b} = \frac{p}{q}\) where \(\frac{p}{q}\) is in lowest terms. Without loss of generality, assume \(p > 0\) and \(q > 0\).

   If \(p > 1\), then \(p\) must divide \(\lfloor a \cdot n \rfloor\) for all \(n\). Choose \(n\) such that \(1 < n \cdot a_1 < p\). Then:
   \[
   \lfloor a \rfloor \cdot n < \lfloor a \cdot n \rfloor < \lfloor a \rfloor \cdot n + p
   \]
   This implies \(\lfloor a \cdot n \rfloor\) cannot be a multiple of \(p\), which is a contradiction. Hence, \(p = 1\).

   Similarly, if \(q > 1\), then \(q\) must divide \(\lfloor b \cdot n \rfloor\) for all \(n\). Choose \(n\) such that \(1 < n \cdot b_1 < q\). Then:
   \[
   \lfloor b \rfloor \cdot n < \lfloor b \cdot n \rfloor < \lfloor b \rfloor \cdot n + q
   \]
   This implies \(\lfloor b \cdot n \rfloor\) cannot be a multiple of \(q\), which is a contradiction. Hence, \(q = 1\).

   Therefore, \(|\frac{a}{b}| = 1 \implies |a| = |b|\).

   - If \(a = -b\), then:
     \[
     a \cdot \lfloor -a \cdot n \rfloor = -a \cdot \lfloor a \cdot n \rfloor
     \]
     For \(n = 1\), this gives:
     \[
     -\lfloor a \rfloor = \lfloor -a \rfloor
     \]
     Since \(a_1 > 0\), \(\lfloor -a \rfloor = -\lfloor a \rfloor - 1\), which is a contradiction.

   - If \(a = b\), the condition is satisfied.

   Thus, the solutions from this case are pairs of the form \((a, a)\), where \(a\) is a real number which is not an integer.

2. **Case 2: \(a\) is an integer different from 0**

   Let \(a\) be an integer. Then:
   \[
   \lfloor a \cdot n \rfloor = a \cdot n
   \]
   This implies:
   \[
   \lfloor b \cdot n \rfloor = b \cdot n
   \]
   For \(n = 1\), we get:
   \[
   \lfloor b \rfloor = b
   \]
   Hence, \(b\) is also an integer. Therefore, the solutions are pairs \((a, b)\), where \(a\) and \(b\) are arbitrary integers.

3. **Case 3: \(a = 0\) or \(b = 0\)**

   If \(a = 0\), then:
   \[
   0 \cdot \lfloor b \cdot n \rfloor = b \cdot \lfloor 0 \cdot n \rfloor = 0
   \]
   This is true for any \(b\). Hence, the solutions are \((0, b)\), where \(b\) is any real number.

   If \(b = 0\), then:
   \[
   a \cdot \lfloor 0 \cdot n \rfloor = 0 \cdot \lfloor a \cdot n \rfloor = 0
   \]
   This is true for any \(a\). Hence, the solutions are \((a, 0)\), where \(a\) is any real number.

The final answer is \( \boxed{ (a, b) } \) such that: