In boxes labeled $0$, $1$, $2$, $\dots$, we place integers according to the following rules:

$\bullet$ If $p$ is a prime number, we place it in box $1$.

$\bullet$ If $a$ is placed in box $m_a$ and $b$ is placed in box $m_b$, then $ab$ is placed in the box labeled $am_b+bm_a$.

Find all positive integers $n$ that are placed in the box labeled $n$.
1. **Base Case:**
   - If \( p \) is a prime number, it is placed in box 1. This is given by the problem statement.

2. **Inductive Step:**
   - Assume that for \( p_1p_2\cdots p_k \), where \( p_1, p_2, \ldots, p_k \) are primes, the number is placed in box \( \left(p_1p_2\cdots p_k\right) \left(\frac{1}{p_1} + \frac{1}{p_2} + \cdots + \frac{1}{p_k}\right) \).
   - Consider the number \( p_1p_2\cdots p_k \cdots p_{k+1} \). According to the rule, it is placed in the box:
     \[
     \left(p_1p_2\cdots p_k\right) \cdot 1 + p_{k+1} \left(p_1p_2\cdots p_k\right) \left(\frac{1}{p_1} + \frac{1}{p_2} + \cdots + \frac{1}{p_k}\right)
     \]
     Simplifying this, we get:
     \[
     \left(p_1p_2\cdots p_{k+1}\right) \cdot \frac{1}{p_{k+1}} + \left(p_1p_2\cdots p_{k+1}\right) \left(\frac{1}{p_1} + \frac{1}{p_2} + \cdots + \frac{1}{p_k}\right)
     \]
     \[
     = \left(p_1p_2\cdots p_{k+1}\right) \left(\frac{1}{p_1} + \frac{1}{p_2} + \cdots + \frac{1}{p_{k+1}}\right)
     \]
     This completes the induction.

3. **Finding \( n \) such that \( n \) is placed in box \( n \):**
   - We need to find \( n = q_1q_2\cdots q_m \) (with \( q_1, q_2, \ldots, q_m \) being primes) such that:
     \[
     \frac{1}{q_1} + \frac{1}{q_2} + \cdots + \frac{1}{q_m} = 1
     \]
   - Multiply both sides by \( q_1q_2\cdots q_m \):
     \[
     q_2q_3\cdots q_m + q_1q_3\cdots q_m + \cdots + q_1q_2\cdots q_{m-1} = q_1q_2\cdots q_m
     \]
   - Consider the equation modulo \( q_1 \):
     \[
     q_2q_3\cdots q_m \equiv 0 \pmod{q_1}
     \]
     This implies that at least one other prime, say \( q_2 \), must be equal to \( q_1 \).

4. **Continuing the process:**
   - Consider the equation modulo \( q_1^2 \):
     \[
     q_1 \cdot (q_3q_4\cdots q_m) + q_1 \cdot (q_3q_4\cdots q_m) \equiv 0 \pmod{q_1^2}
     \]
     This implies:
     \[
     2q_3q_4\cdots q_m \equiv 0 \pmod{q_1}
     \]
     So either \( q_1 = 2 \) or \( q_3q_4\cdots q_m \equiv 0 \pmod{q_1} \). This process continues, showing that all primes must be equal.

5. **Conclusion:**
   - All primes must be equal, so \( n = p^p \) where \( p \) is a prime number.

The final answer is \( \boxed{ n = p^p } \) where \( p \) is a prime number.