Consider $ x_1, x_2, ..., x_n \in \mathbb{R}$.
We have to prove that each $\sqrt x $ is rational if the sum of $\sqrt x_1 + \ldots + \sqrt x_n $ is rational.
I think that I could prove it using fact, that only the sum of the opposite irrational number is rational for example $ \sqrt 2 + (2-\sqrt 3) = 2 $, because if $ x_1, x_2, ..., x_n \in \mathbb{R} $ then $ \sqrt (-1) \neq - \sqrt 1$.
On
As stated, with real numbers $x_i$, it's false. After all, we could take something like $x_1=(\sqrt{2})^2$, $x_2=(2-\sqrt{2})^2$. For this to make sense, those $x_i$ must all be rational.
For $n=2$, this is easy to do in an elementary way. We can solve for one of the square roots and then square: \begin{align*}x_1+x_2 &= S\\ \sqrt{x_1} &= S - \sqrt{x_2}\\ x_1 &= S^2 + x_2 - 2S\sqrt{x_2}\\ \sqrt{x_2} &= \frac{S^2+x_2-x_1}{2S}\end{align*} With one of the square roots rational, the other has to be as well.
This does not generalize well, at least at this level of understanding. With larger numbers of square roots involved, squaring the other side leads to the square roots of the products getting involved, increasing the total number of square roots involved. We can work through it, but it requires the language of linear algebra and field extensions to make sense of it in general.
See also here for a proof of the general case.
On
As mentioned there are simple counterexamples. However the following version is true
Theorem $\rm\ \sqrt{c_1}+\cdots+\!\sqrt{c_{n}} \in K\ \Rightarrow \sqrt{c_i}\in K\:$ for all $\rm i,\:$ if $\rm\: 0 < c_i\in K$ an ordered field.
Proof $ $ See this answer for a simple inductive proof.
Let $$x_1=2$$ $$x_2=(2-\sqrt{2})^2=6-4\sqrt{2}$$ Then $$\sqrt{x_1}+\sqrt{x_2}=2$$ But $$x_2\not\in \mathbb{Q}$$