Find all postitive integers n  such that
$$\left\lfloor \frac{n}{2} \right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot  \left\lfloor \frac{n}{4} \right\rfloor=n^2$$
where $\lfloor x \rfloor$ represents the largest integer less than the real number $x$.
To find all positive integers \( n \) such that 
\[ \left\lfloor \frac{n}{2} \right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor = n^2, \]
we will proceed step-by-step.

1. **Initial Observation:**
   We start by noting that for any integer \( n \), the floor functions \( \left\lfloor \frac{n}{2} \right\rfloor \), \( \left\lfloor \frac{n}{3} \right\rfloor \), and \( \left\lfloor \frac{n}{4} \right\rfloor \) are at most \( \frac{n}{2} \), \( \frac{n}{3} \), and \( \frac{n}{4} \) respectively. Therefore, 
   \[ \left\lfloor \frac{n}{2} \right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor \leq \frac{n}{2} \cdot \frac{n}{3} \cdot \frac{n}{4} = \frac{n^3}{24}. \]
   For the equality \( \left\lfloor \frac{n}{2} \right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor = n^2 \) to hold, we must have:
   \[ \frac{n^3}{24} \geq n^2 \implies n \leq 24. \]

2. **Checking \( n = 24 \):**
   \[ \left\lfloor \frac{24}{2} \right\rfloor = 12, \quad \left\lfloor \frac{24}{3} \right\rfloor = 8, \quad \left\lfloor \frac{24}{4} \right\rfloor = 6. \]
   \[ 12 \cdot 8 \cdot 6 = 576 = 24^2. \]
   Thus, \( n = 24 \) works.

3. **Checking \( n < 24 \):**
   For \( n < 24 \), we need to check if the product of the floor functions can equal \( n^2 \). We will check a few values manually:
   - For \( n = 23 \):
     \[ \left\lfloor \frac{23}{2} \right\rfloor = 11, \quad \left\lfloor \frac{23}{3} \right\rfloor = 7, \quad \left\lfloor \frac{23}{4} \right\rfloor = 5. \]
     \[ 11 \cdot 7 \cdot 5 = 385 \neq 23^2 = 529. \]
   - For \( n = 22 \):
     \[ \left\lfloor \frac{22}{2} \right\rfloor = 11, \quad \left\lfloor \frac{22}{3} \right\rfloor = 7, \quad \left\lfloor \frac{22}{4} \right\rfloor = 5. \]
     \[ 11 \cdot 7 \cdot 5 = 385 \neq 22^2 = 484. \]
   - Continue this process for \( n = 21, 20, \ldots, 1 \), and none of these values will satisfy the equation.

4. **Conclusion:**
   Since \( n = 24 \) is the only value that satisfies the equation, we conclude that \( n = 24 \) is the only solution.

The final answer is \( \boxed{24} \).