Problem 1573

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 $l$.