To determine if Baron Munchausen's theorem is correct, we need to show that for any positive integers $a$ and $b$, there exists a positive integer $n$ such that $an$ is a perfect square and $bn$ is a perfect cube.

1. Prime Factorization:
Let $a$ and $b$ be expressed in their prime factorizations:
$$a = p_1^{x_1} p_2^{x_2} \ldots p_k^{x_k}$$
$$b = p_1^{y_1} p_2^{y_2} \ldots p_k^{y_k}$$
where $x_i, y_i \geq 0$ for all $i$.

2. **Form of $n$**:
Assume $n$ can be written as:
$$n = p_1^{z_1} p_2^{z_2} \ldots p_k^{z_k}$$
where $z_i \geq 0$ for all $i$.

3. **Conditions for $an$ to be a Perfect Square**:
For $an$ to be a perfect square, the exponent of each prime in its factorization must be even. Therefore, we need:
$$x_i + z_i \equiv 0 \pmod{2} \quad \text{for all } i$$

4. **Conditions for $bn$ to be a Perfect Cube**:
For $bn$ to be a perfect cube, the exponent of each prime in its factorization must be divisible by 3. Therefore, we need:
$$y_i + z_i \equiv 0 \pmod{3} \quad \text{for all } i$$

5. Chinese Remainder Theorem:
We need to find $z_i$ such that both conditions are satisfied simultaneously. This can be formulated as a system of congruences:
$$z_i \equiv -x_i \pmod{2}$$
$$z_i \equiv -y_i \pmod{3}$$

By the Chinese Remainder Theorem, since 2 and 3 are coprime, there exists a unique solution modulo 6 for each $z_i$. Therefore, there exists $z_i$ such that:
$$z_i \equiv -x_i \pmod{2}$$
$$z_i \equiv -y_i \pmod{3}$$

6. Conclusion:
Since we can find such $z_i$ for each prime factor, we can construct $n$ such that $an$ is a perfect square and $bn$ is a perfect cube.

$\blacksquare$