Problem 1311

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 $\boxed{ n \ge 4 }$.