I have been organizing my language, if there is anything not clear about the question, I am sorry.
The question should be:
$\forall n\in\mathbb N^*, a_i\in\mathbb N$, $a_i$ is not a perfect square for $1\leq i\leq n$. Then $\sum_{i=1}^n \sqrt{a_i}$ is not a rational number.
(We assume every real number has a square root, and 0 is a perfect square.)
I saw the proof on the website as $n=2$, it is great, and very clever. And it can even prove $n=3$, this statement is true.
However, I think the key in the proof is to change all the square roots into rational numbers, so we get a contradiction. Yet when $n$ is bigger than $3$, square both sides will not be possible to reduce the number of square roots (at least I think). Thus it may not be effective from here.
I think the statement is true, because I think two positive irrational numbers sum up and get rational numbers will only be in form like $2-\sqrt 2$ and $\sqrt 2$, and something like $2-\sqrt 2$ cannot be $\sqrt n$.
However, I cannot think of a way to prove it. This reminds me that when I want to prove something is a irrational number, all the way I have is to assume it is not, and get contradiction. I think this is not good, so if you have some other methods to prove something irrational, and willing to share with me, I will really be appreciated.