Given a positive integer $N$, determine all positive integers $n$, satisfying the following condition: for any list $d_1,d_2,\ldots,d_k$ of (not necessarily distinct) divisors of $n$ such that $\frac{1}{d_1} + \frac{1}{d_2} + \ldots + \frac{1}{d_k} > N$, some of the fractions $\frac{1}{d_1}, \frac{1}{d_2}, \ldots, \frac{1}{d_k}$ add up to exactly $N$.
1. **Identify the problem and the given condition:**
   We need to determine all positive integers \( n \) such that for any list \( d_1, d_2, \ldots, d_k \) of (not necessarily distinct) divisors of \( n \) satisfying 
   \[
   \frac{1}{d_1} + \frac{1}{d_2} + \ldots + \frac{1}{d_k} > N,
   \]
   some of the fractions \( \frac{1}{d_1}, \frac{1}{d_2}, \ldots, \frac{1}{d_k} \) add up to exactly \( N \).

2. **Propose the solution:**
   The solution claims that the answer is all powers of primes, i.e., \( n = p^l \) for some prime \( p \) and positive integer \( l \).

3. **Verify the solution for powers of primes:**
   Consider \( n = p^l \). Let \( d_i = p^{a_i} \) for \( 1 \leq i \leq k \). Assume that 
   \[
   \sum_{i=1}^k \frac{1}{p^{a_i}} > N.
   \]
   Let \( t = \max(a_1, \ldots, a_k) \). Then, we can rewrite the inequality as:
   \[
   \sum_{i=1}^k \frac{1}{p^{a_i}} = \sum_{i=1}^k \frac{p^{t-a_i}}{p^t} = \frac{1}{p^t} \sum_{i=1}^k p^{t-a_i} > N.
   \]
   This implies:
   \[
   \sum_{i=1}^k p^{t-a_i} > N p^t.
   \]
   We need to find a subset \( S \subseteq \{1, 2, \ldots, k\} \) such that:
   \[
   \sum_{i \in S} p^{t-a_i} = N p^t.
   \]

4. **Prove the claim:**
   Let \( x_1, \ldots, x_k \) be positive integers less than \( t \) such that:
   \[
   \sum_{i=1}^k p^{x_i} > N p^t.
   \]
   We need to show that there exists a subset \( S \subseteq \{1, 2, \ldots, k\} \) such that:
   \[
   \sum_{i \in S} p^{x_i} = N p^t.
   \]
   We use induction on \( t \):
   - **Base case \( t = 1 \):** Clear since \( p^{x_i} = 1 \) and we can always find such a subset.
   - **Inductive step:** Assume the statement holds for \( t-1 \). If all \( x_i \) are larger than 1, divide the inequality by \( p \) and reduce to the case for \( t-1 \). If some \( x_i = 1 \), let \( r \) be the number of such \( x_i \). Then:
     \[
     r + \sum_{1 \leq i \leq k, x_i \neq 1} p^{x_i} > N p^t.
     \]
     If \( \sum_{i=1}^k p^{x_i} = N p^t + u \) where \( u \) is a positive integer, we have two cases:
     - If \( r \geq u \), pick all \( x_i \neq 1 \) and \( r-u \) of the \( x_i = 1 \).
     - If \( r < u \), then \( \sum_{1 \leq i \leq k, x_i \neq 1} p^{x_i} > N p^t \) and reduce to the case of \( t-1 \).

5. **Conclude the proof for powers of primes:**
   The claim ensures that we can choose a subset \( S \) such that:
   \[
   \sum_{i \in S} p^{a_i} = N p^t.
   \]
   This proves that all powers of primes indeed work.

6. **Disprove other cases:**
   Assume \( n \) has two prime divisors \( p < q \). Consider:
   \[
   \sum_{i=1}^{q-1} \frac{1}{q} + \frac{1}{p} > 1.
   \]
   There must exist a positive integer \( k \) such that:
   \[
   \frac{k}{q} + \frac{1}{p} = 1,
   \]
   which is impossible. This contradiction shows that \( n \) cannot have more than one prime divisor.

The final answer is \( \boxed{ n = p^l } \) for some prime \( p \) and positive integer \( l \).