Prove that for all naturals $n \ge 6$ there is a set of $n$ positive naturals, $a_1$ to $a_n$ such that $\sum_{i=1}^n \left(\frac{1}{a_i}\right)^2 =1$ I don't know how to prove this. I know that $\{2, 2, 2, 2\}$ is a set for $n = 4$, since $\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2  = 1$. So a set of $n$ naturals can have repeated elements.

Extending your observation, you can replace $a$ in any list with $2a, 2a, 2a, 2a$, to get a new valid list. Hence if it is possible for $n$ then it is possible for $n + 3$. Then you only need to prove it for $6, 7, 8$.
Another nice fact:
$$\frac{1}{4} = \frac{1}{9} + \frac{1}{9} + \frac{1}{36}$$
So if $2$ is in your list, you can replace it with $3, 3, 6$.
Starting with $(2,2,2,2)$, applying the first strategy gives $(2,2,2,4,4,4,4)$, solving $n = 7$.
Applying the second strategy gives $(2,2,2,3,3,6)$, solving $n = 6$. And applying the second strategy again gives $(2,2,3,3,3,3,6)$, solving $n = 8$.
