Answer: All naturals bigger than $3$. We are going to construct inductively, For $n=4$ take $(a_1,a_2,a_3,a_4)=(1,2,3,6)$ For $n=5$ take $(a_1,a_2,a_3,a_4,a_5)=(9,25,30,36,100)$ Use the identity below for the induction step: $$ \frac{1}{x}=\frac{1}{x+2}+\frac{1}{x^2+x}+\frac{1}{x^2+3x+2} $$thus we can get a construction for $n+2$ from the construction of $n$. It's trivial to see why $n=3$ does not works.

Remark

I proved even case and for all $n$ my inductive step was like this: $\frac{1}{x}=\frac{1}{2x}+\frac{1}{3x}+\frac{1}{6x}$. But I got stuck on finding an example for $n=5$. Can you please share your motivation behind this example?

Jalil_Huseynov wrote: I proved even case and for all $n$ my inductive step was like this: $\frac{1}{x}=\frac{1}{2x}+\frac{1}{3x}+\frac{1}{6x}$. But I got stuck on finding an example for $n=5$. Can you please share your motivation behind this example? The inductive step you mentioned is indeed correct. I first noticed an alternative formulation: if $a_n$ is a perfect number with $n$ divisors $a_1, \cdots, a_n$ then the sequence holds. Thus when $n=2p$ we may take $a_n = 2^{p-1}(2^p-1)$. Generalizing this to all $n$ even and only choosing a suitable subset of its divisors, we get \[a_1, \cdots, a_{n/2} = 1, 2, 4, \cdots, 2^{n/2-1}\qquad a_{n/2+1}, \cdots, a_n = (2^{n/2} - 1), 2(2^{n/2} - 1),\cdots, 2^{n/2-1}(2^{n/2} - 1) \] For $n=5$, the key is to find a sequence with $a_1a_5=a_2a_4=a_3^2$ so that we only need to worry about one of the equalities. I forgot the details but I ended up trying something like $1, x, xy, xy^2, x^2y^2$ for some rationals $x > 1, y > 1$, find a working example, and then scale to integer.

The case $n=5$ has proved to be difficult (I kinda underestimated its difficulty, unfortunately), so here I will include some motivations on how to find the solution above. First note that it is possible to relax the condition to $a_i$ being rationals, as both equations are homogeneous, hence invariant up to scaling. Next, to eliminate one of the equations, it is natural to want $a_2$, $a_4$ and $a_3$ to behave symmetrically, that is $a_2a_4=a_3^2=a_5a_1$ so that $$\frac{1}{a_1}=\frac{1}{a_2}+...+\frac{1}{a_n} \iff a_5=\frac{a_1a_5}{a_2}+\frac{a_5a_1}{a_3}+\frac{a_5a_1}{a_4}+a_1=a_4+a_3+a_2+a_1$$Such choices of $(a_1,a_2,\cdots,a_5)$ can be parameterized by $\left(1, d, n, \displaystyle \frac{n^2}{d}, n^2\right)$ (with $n, d>1$ are rationals). The equation then becomes $$n^2\left(1-\frac{1}{d}\right)-n-(d+1)=0$$which has rational solutions $x$ if the discriminant $$\Delta=1+4\left(1-\frac1d\right)(d+1)$$is a rational square for some rational $d$. This turns out to have an integer solution: multiplying by $d^2$, then it can be factored as $d(4d^2+d-4)$, so choosing $d=4$ gives a perfect square. This corresponds to $n=\frac{10}{3}$ and thus we get the solution $(1,4,\frac{20}{6},\frac{100}{36},\frac{100}{9})$, which upon rearranging the terms and scaling appropriately gives $(9, 25, 30, 36, 100)$ as we wanted.

Official Solution: Answer: All $n\ge 4$. Solution: For $n=3$ this is impossible, because if $a<b<c$ are integers such that $c=a+b$ and $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$, then we can rewrite it as $$\frac{1}{a}=\frac{1}{b}+\frac{1}{a+b}$$$$\iff b^2-ab-a^2=0$$This implies the discriminant $\Delta=a^2-4(-a^2)=5a^2$ is a perfect square, which is impossible. For $n=4$ and $n=5$ we have the solutions $(1,2,3,6)$ and $(9, 25, 30, 36, 100)$ respectively. To obtain solutions for $n\ge 6$, we proceed by induction on $n$. If we have $(a_1, \cdots, a_n)$ as a solution for $n$, then we can construct $(a_1, \cdots, a_{n-1}, 2a_n, 3a_n, 6a_n)$ for $n+2$, this works because $$a_n=a_1+...+a_{n-1} \Rightarrow 6a_n=a_1+...+a_{n-1}+2a_n+3a_n$$Likewise we have $$\frac{1}{a_1}=\frac{1}{a_2}+...+\frac{1}{a_n} \Rightarrow \frac{1}{a_1}=\frac{1}{a_2}+...+\frac{1}{a_{n-1}}+\frac{1}{2a_n}+\frac{1}{3a_n}+\frac{1}{6a_n}$$which completes the induction from $n$ to $n+2$. So all $n\ge 4$ are possible. $\blacksquare$ Remark: A construction for $n$ even case that bypasses the induction protocol is possible. Indeed, consider the $n=2m$ numbers where $a_1, \cdots, a_{m} = 1, 2, \cdots, 2^{m - 1}$ and $a_{m + 1}, \cdots, a_{2m} = 2^m - 1, 2\cdot (2^m - 1), \cdots, 2^{m - 1}(2^m - 1)$. This construction is motivated by perfect numbers in the form $2^{p-1}(2^p-1)$ where $p$ is prime. Using this, construction for $n=5$ can now be motivated by finding examples in the form $\{1, x, xy, xy^2, x^2y^2\}$ (with $x, y > 1$ rational). Some trial and error yields that we can take $x = \left(\frac 53\right)^2$ and $y = \frac 65$. [The thought process of finding this pair can also be refered to the post/remark above]