Determine all pairs $(a,b)$ of real numbers such that $a \lfloor bn \rfloor =b \lfloor an \rfloor $ for all positive integers $n$. (Note that $\lfloor x\rfloor $ denotes the greatest integer less than or equal to $x$.)
To determine all pairs \((a, b)\) of real numbers such that \(a \lfloor bn \rfloor = b \lfloor an \rfloor\) for all positive integers \(n\), we will analyze different cases for \(a\) and \(b\).

1. **Case 1: \(a = b\)**
   - If \(a = b\), then the equation becomes \(a \lfloor an \rfloor = a \lfloor an \rfloor\), which is always true. Therefore, any pair \((a, a)\) satisfies the condition.

2. **Case 2: \(a = 0\) or \(b = 0\)**
   - If \(a = 0\), the equation becomes \(0 = b \lfloor 0 \rfloor = 0\), which is true for any \(b\).
   - If \(b = 0\), the equation becomes \(a \lfloor 0 \rfloor = 0 = 0\), which is true for any \(a\).
   - Therefore, any pair \((0, b)\) or \((a, 0)\) satisfies the condition.

3. **Case 3: \(a \neq b\) and \(a, b \neq 0\)**
   - Assume \(a\) and \(b\) are both non-zero and \(a \neq b\). We need to show that this case fails.

4. **Subcase 3.1: \(a\) is irrational and \(b\) is rational**
   - If \(a\) is irrational and \(b\) is rational, then \(\lfloor an \rfloor\) will not have a simple relationship with \(\lfloor bn \rfloor\). This will generally fail to satisfy the equation \(a \lfloor bn \rfloor = b \lfloor an \rfloor\).

5. **Subcase 3.2: Both \(a\) and \(b\) are irrational**
   - If both \(a\) and \(b\) are irrational, and \(a/b\) is irrational, then the equation will not hold for all \(n\) because the floors of irrational multiples will not align in a way that satisfies the equation.

6. **Subcase 3.3: Both \(a\) and \(b\) are rational**
   - Let \(a = \frac{p}{q}\) and \(b = \frac{r}{s}\) where \(p, q, r, s\) are integers and \(\gcd(p, q) = 1\) and \(\gcd(r, s) = 1\).
   - We need to satisfy:
     \[
     \frac{p}{q} \lfloor \frac{rn}{s} \rfloor = \frac{r}{s} \lfloor \frac{pn}{q} \rfloor
     \]
   - This implies:
     \[
     ps \lfloor \frac{rn}{s} \rfloor = rq \lfloor \frac{pn}{q} \rfloor
     \]
   - For this to hold for all \(n\), the fractional parts must align such that:
     \[
     ps \left\{ \frac{rn}{s} \right\} = rq \left\{ \frac{pn}{q} \right\}
     \]
   - Given \(\gcd(q, s) = 1\), the only way this can hold is if \(s \mid r\) and \(q \mid p\), which would imply \(a = b\), contradicting our assumption that \(a \neq b\).

Therefore, the only solutions are when \(a = b\), \(a = 0\), or \(b = 0\).

The final answer is \( \boxed{ (a, b) = (c, c) } \) for any real number \(c\), or \((a, 0)\) for any real number \(a\), or \((0, b)\) for any real number \(b\).