Problem 1412

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 $\boxed{24}$.