Determine all squarefree positive integers $n\geq 2$ such that \[\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_k}\]is a positive integer, where $d_1,d_2,\ldots,d_k$ are all the positive divisors of $n$.
To determine all squarefree positive integers \( n \geq 2 \) such that 
\[
\frac{1}{d_1} + \frac{1}{d_2} + \cdots + \frac{1}{d_k}
\]
is a positive integer, where \( d_1, d_2, \ldots, d_k \) are all the positive divisors of \( n \), we proceed as follows:

1. **Express the sum of reciprocals of divisors**:
   If \( d_i \) is a divisor of \( n \), then so is \( \frac{n}{d_i} \). Therefore, we can pair each \( d_i \) with \( \frac{n}{d_i} \). This gives us:
   \[
   \sum_{i=1}^k \frac{1}{d_i} = \sum_{i=1}^k \frac{d_i}{n} = \frac{d_1 + d_2 + \cdots + d_k}{n}
   \]
   where \( \sigma(n) = d_1 + d_2 + \cdots + d_k \) is the sum of the divisors of \( n \).

2. **Condition for the sum to be an integer**:
   For the sum to be an integer, \( \frac{\sigma(n)}{n} \) must be an integer. This implies that \( n \) must divide \( \sigma(n) \).

3. **Properties of squarefree numbers**:
   Since \( n \) is squarefree, it can be written as \( n = p_1 p_2 \cdots p_m \) where \( p_i \) are distinct primes. The sum of the divisors of \( n \) is given by:
   \[
   \sigma(n) = (1 + p_1)(1 + p_2) \cdots (1 + p_m)
   \]

4. **Analyzing the divisibility**:
   We need \( n \leq \sigma(n) \). Note that \( \sigma(n) \) is always even if \( n \) is even. Therefore, \( n \) must be even. Let:
   \[
   n = 2 \cdot p_2 \cdot \ldots \cdot p_m
   \]
   Then:
   \[
   \sigma(n) = (1 + 2)(1 + p_2)(1 + p_3) \cdots (1 + p_m)
   \]

5. **Considering the largest prime divisor**:
   Let \( p_m \) be the greatest prime divisor of \( n \) other than 2. Then \( p_m \leq p_j + 1 \) for some \( j \). Since \( p_m \) is the greatest prime, there cannot be two neighboring primes other than 2 and 3. This leads to a contradiction unless the only primes are 2 and 3.

6. **Conclusion**:
   The only possible \( n \) is \( n = 6 \) (since 6 is the product of the primes 2 and 3).

The final answer is \( \boxed{ n = 6 } \).