Let $f_0, f_1,...$ be the Fibonacci sequence: $f_0 = f_1 = 1, f_n = f_{n-1} + f_{n-2}$ if $n \geq 2$.
Determine all possible positive integers $n$ so that there is a positive integer $a$ such that
$f_n \leq a \leq f_{n+1}$ and that $a(  \frac{1}{f_1}+\frac{1}{f_1f_2}+\cdots+\frac{1}{f_1f_2...f_n}  )$
is an integer.
1. **Define the Fibonacci sequence**: The Fibonacci sequence is defined as follows:
   \[
   f_0 = f_1 = 1, \quad f_n = f_{n-1} + f_{n-2} \quad \text{for} \quad n \geq 2
   \]

2. **Given condition**: We need to find all possible positive integers \( n \) such that there exists a positive integer \( a \) satisfying:
   \[
   f_n \leq a \leq f_{n+1}
   \]
   and
   \[
   a \left( \frac{1}{f_1} + \frac{1}{f_1 f_2} + \cdots + \frac{1}{f_1 f_2 \cdots f_n} \right)
   \]
   is an integer.

3. **Simplify the expression**: The given expression can be expanded as:
   \[
   a \left( \frac{1}{f_1} + \frac{1}{f_1 f_2} + \cdots + \frac{1}{f_1 f_2 \cdots f_n} \right)
   \]
   This can be written as:
   \[
   a \left( \frac{f_2 f_3 \cdots f_n + f_3 \cdots f_n + \cdots + f_{n-1} f_n + f_n + 1}{f_1 f_2 \cdots f_n} \right)
   \]

4. **Divisibility condition**: For the above expression to be an integer, the numerator must be divisible by the denominator. This implies:
   \[
   f_n \mid a \left( f_2 f_3 \cdots f_n + f_3 \cdots f_n + \cdots + f_{n-1} f_n + f_n + 1 \right)
   \]
   Since \( f_n \) is a term in the Fibonacci sequence, it must divide \( a \). Therefore, \( a \) must be a multiple of \( f_n \).

5. **Range of \( a \)**: Given \( f_n \leq a \leq f_{n+1} \), and since \( a \) must be a multiple of \( f_n \), the only possible value for \( a \) is \( f_n \) itself.

6. **Check the condition for \( a = f_n \)**: We need to verify if:
   \[
   f_n \left( \frac{1}{f_1} + \frac{1}{f_1 f_2} + \cdots + \frac{1}{f_1 f_2 \cdots f_n} \right)
   \]
   is an integer. This simplifies to:
   \[
   \frac{f_n (f_2 f_3 \cdots f_n + f_3 \cdots f_n + \cdots + f_{n-1} f_n + f_n + 1)}{f_1 f_2 \cdots f_n}
   \]
   which further simplifies to:
   \[
   \frac{f_2 f_3 \cdots f_n + f_3 \cdots f_n + \cdots + f_{n-1} f_n + f_n + 1}{f_1 f_2 \cdots f_{n-1}}
   \]

7. **Inductive proof**: We use induction to prove that \( f_{n-1} \mid f_n + 1 \) for \( n \geq 3 \). The base cases for \( n = 2, 3 \) are true. Assume it holds for \( n = k \), then for \( n = k+1 \):
   \[
   f_{k+1}^2 - f_k f_{k+2} = f_{k+1}^2 - f_k (f_{k+1} + f_k) = f_{k+1} (f_{k+1} - f_k) - f_k^2 = f_{k+1} f_{k-1} - f_k^2 = -(-1)^k = (-1)^{k+1}
   \]
   This completes the inductive step.

8. **Conclusion**: Using the claim, we find that \( n \) must be even. Additionally, we use the fact that \( f_{n-2} \mid f_{n-1} f_n + f_n + 1 \) and the properties of Fibonacci numbers to conclude that \( f_{n-2} \mid f_{n-1} \), which is only possible for small values of \( n \).

9. **Final cases**: We check the cases \( n = 1, 2, 3 \):
   - For \( n = 1 \): \( f_1 = 1, f_2 = 1 \), so \( a \in \{1, 2\} \).
   - For \( n = 2 \): \( f_2 = 1, f_3 = 2 \), so \( a = 2 \).
   - For \( n = 3 \): \( f_3 = 2, f_4 = 3 \), so \( a = 3 \).

The final answer is \( \boxed{ n \in \{1, 2, 3\} } \) and corresponding \( a \in \{1, 2, 3\} \).