Find, with proof, all positive integers $n$ with the following property: There are only finitely many positive multiples of $n$ which have exactly $n$ positive divisors
1. **Define the problem and notation:**
   - We need to find all positive integers \( n \) such that there are only finitely many positive multiples of \( n \) which have exactly \( n \) positive divisors.
   - Let \(\sigma(n)\) denote the number of positive divisors of \( n \).
   - Let \(\mathbb{P}\) be the set of all primes.
   - Let \( D_n \) denote the set of prime divisors of \( n \).

2. **Lemma 1:**
   - For any positive integers \( n \) and \( k \), \(\sigma(kn) \geq \sigma(n)\), with equality occurring only when \( k = 1 \).
   - **Proof:**
     - If \( k = 1 \), the lemma is trivially true.
     - Suppose \( k > 1 \). Any divisor of \( n \) must also be a divisor of \( kn \), but \( kn \) itself is a divisor of \( kn \) that does not divide \( n \). Thus, \(\sigma(kn) > \sigma(n)\) for \( k > 1 \).
     - \(\blacksquare\)

3. **Lemma 2:**
   - Given fixed nonnegative integers \( x_1, x_2, \ldots, x_k \) and a positive integer \( n \), there are finitely many ordered tuples of positive integers \( (y_1, y_2, \ldots, y_k) \) which satisfy \(\prod_{i=1}^k (x_i + y_i) = n\).
   - **Proof:**
     - Since everything being multiplied is a positive integer, we can establish the upper bound \( y_i \leq n \) for all \( i \), giving an upper bound of \( n^k \) possible ordered tuples.
     - Since \( n^k \) is finite, the number of such tuples is finite.
     - \(\blacksquare\)

4. **Lemma 3:**
   - The given condition that there are only finitely many positive multiples of \( n \) which have exactly \( n \) positive divisors is satisfied if and only if there does not exist a positive integer \( k \) such that \(\sigma(kn) \mid n\) and \(\sigma(kn) \neq n\).
   - **Proof:**
     - **If part:**
       - Suppose there exists a positive integer \( k \) such that \(\sigma(kn) \mid n\) and \(\sigma(kn) \neq n\).
       - Let \( m = \frac{n}{\sigma(kn)} \). Then \( m \) is a positive integer at least 2.
       - Every number of the form \( knp^{m-1} \) where \( p \in \mathbb{P} \setminus D_{kn} \) has \( n \) divisors.
       - Since \( |\mathbb{P}| \) is infinite and \( |D_{kn}| \) is finite, there are infinitely many such multiples.
       - Thus, if such a \( k \) exists, the condition is not satisfied.
     - **Only if part:**
       - Suppose there does not exist a positive integer \( k \) such that \(\sigma(kn) \mid n\) and \(\sigma(kn) \neq n\).
       - Any positive integer \( k \) such that \(\sigma(kn) = n\) must satisfy \( D_k \subseteq D_n \).
       - Otherwise, there would exist a prime \( q \in D_k \setminus D_n \), leading to a contradiction.
       - Thus, any \( k \) such that \(\sigma(kn) = n\) satisfies \( D_{kn} = D_n \).
       - Using the formula for \(\sigma(kn)\), there are finitely many ordered tuples \((b_1, b_2, \ldots, b_m)\) such that \(\prod_{i=1}^m (a_i + b_i + 1) = n\), by Lemma 2.
       - \(\blacksquare\)

5. **Positive integers not divisible by the square of any prime:**
   - For \( n = 1 \), the only positive integer with one divisor is 1.
   - For \( n \neq 1 \), write \( n = \prod_{i=1}^m p_i \) for distinct primes \( p_i \).
   - Any \( k \) such that \(\sigma(kn) = n\) must satisfy \( D_k \subseteq D_n \).
   - Using the formula for \(\sigma(kn)\), there are finitely many ordered tuples \((a_1, a_2, \ldots, a_m)\) such that \(\prod_{i=1}^m (a_i + 1) = n\), by Lemma 2.
   - Thus, any positive integers not divisible by the square of any prime work.

6. **Special case for \( n = 4 \):**
   - \(\sigma(4) = 3\), so for any positive integer \( k \), \(\sigma(4k) \geq 3\).
   - Thus, there does not exist a positive integer \( k \) such that \(\sigma(4k) \mid 4\) and \(\sigma(4k) \neq 4\).
   - By Lemma 3, there are only finitely many positive multiples of 4 which have exactly 4 divisors.

7. **Induction for other cases:**
   - **Base cases:**
     - For \( 2^k \) with \( k \geq 3 \), \(\sigma(2^k) = k + 1 \leq 2^{k-1}\).
     - For \( p^k \) with \( k \geq 2 \) and \( p \in \mathbb{P} \setminus \{2\} \), \(\sigma(p^k) = k + 1 \leq p^{k-1}\).
   - **Inductive step:**
     - Suppose \( n = \prod_{i=1}^m p_i^{a_i} \) doesn't work.
     - There exists a positive integer \( k \) such that \(\sigma(kn) \mid n\) and \(\sigma(kn) \neq n\).
     - Multiplying \( n \) by \( q^b \) for \( q \in \mathbb{P} \setminus D_n \) and \( b \in \mathbb{Z}^+ \) leads to infinitely many multiples with \( n \cdot q^b \) divisors.
     - This completes the inductive step.

The final answer is all positive integers which are not divisible by the square of any prime, as well as \( \boxed{ 4 } \).