For which $n\ge 3$ does there exist positive integers $a_1<a_2<\cdots <a_n$, such that: $$a_n=a_1+...+a_{n-1}, \hspace{0.5cm} \frac{1}{a_1}=\frac{1}{a_2}+...+\frac{1}{a_n}$$ are both true?

[i]Proposed by Ivan Chan[/i]
1. **Case \( n = 3 \):**
   - Suppose \( a_1 < a_2 < a_3 \) are positive integers such that \( a_3 = a_1 + a_2 \) and \( \frac{1}{a_1} = \frac{1}{a_2} + \frac{1}{a_3} \).
   - Substituting \( a_3 = a_1 + a_2 \) into the second equation, we get:
     \[
     \frac{1}{a_1} = \frac{1}{a_2} + \frac{1}{a_1 + a_2}
     \]
   - Clearing the denominators by multiplying through by \( a_1 a_2 (a_1 + a_2) \), we obtain:
     \[
     a_2 (a_1 + a_2) = a_1 (a_1 + a_2) + a_1 a_2
     \]
   - Simplifying, we get:
     \[
     a_2 a_1 + a_2^2 = a_1^2 + a_1 a_2 + a_1 a_2
     \]
     \[
     a_2^2 = a_1^2 + a_1 a_2
     \]
   - Rearranging, we have:
     \[
     a_2^2 - a_1 a_2 - a_1^2 = 0
     \]
   - This is a quadratic equation in \( a_2 \). The discriminant of this quadratic equation must be a perfect square for \( a_2 \) to be an integer:
     \[
     \Delta = (a_1)^2 + 4(a_1^2) = 5a_1^2
     \]
   - Since \( 5a_1^2 \) is not a perfect square for any integer \( a_1 \), there are no solutions for \( n = 3 \).

2. **Case \( n = 4 \):**
   - Consider the sequence \( (1, 2, 3, 6) \):
     - \( a_4 = 6 = 1 + 2 + 3 \)
     - \( \frac{1}{1} = 1 = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \)
   - This sequence satisfies both conditions.

3. **Case \( n = 5 \):**
   - Consider the sequence \( (9, 25, 30, 36, 100) \):
     - \( a_5 = 100 = 9 + 25 + 30 + 36 \)
     - \( \frac{1}{9} = \frac{1}{25} + \frac{1}{30} + \frac{1}{36} + \frac{1}{100} \)
   - This sequence satisfies both conditions.

4. **Induction for \( n \ge 6 \):**
   - Assume we have a solution for \( n \) given by \( (a_1, a_2, \ldots, a_n) \).
   - Construct a solution for \( n + 2 \) as \( (a_1, a_2, \ldots, a_{n-1}, 2a_n, 3a_n, 6a_n) \):
     - The sum condition:
       \[
       a_n = a_1 + a_2 + \cdots + a_{n-1} \implies 6a_n = a_1 + a_2 + \cdots + a_{n-1} + 2a_n + 3a_n
       \]
     - The reciprocal condition:
       \[
       \frac{1}{a_1} = \frac{1}{a_2} + \cdots + \frac{1}{a_n} \implies \frac{1}{a_1} = \frac{1}{a_2} + \cdots + \frac{1}{a_{n-1}} + \frac{1}{2a_n} + \frac{1}{3a_n} + \frac{1}{6a_n}
       \]
   - This completes the induction step, showing that if a solution exists for \( n \), a solution also exists for \( n + 2 \).

5. **Conclusion:**
   - For \( n = 3 \), no solution exists.
   - For \( n \ge 4 \), solutions exist as shown by specific examples and the induction step.

The final answer is \( \boxed{ n \ge 4 } \).