Among $2n-1$ irrationals there are $x_1,\dots,x_n$ such that for rationals $a_i\ge0$ with $\sum_{i=1}^{n}a_i>0$, $\sum_{i=1}^{n}a_ix_i$ is irrational Given a set $S=\{z_1,\cdots, z_{2n-1}\}\subset \Bbb Q^c,$ of $2n-1$ different irrational numbers, prove that there are $n$ different elements $x_1,\cdots,x_n\in S$ such that for all non-negative rational numbers $a_1,\cdots,a_n\in\Bbb Q$, with $a_1+\cdots+a_n>0 $ we have that 
$$a_1x_1+\cdots+a_nx_n$$ is an irrational number. 
I have checked that this is true for $n=1,2,3,4$, but could not figure out how to use induction to argue in general.
Any thoughts are greatly appreciated.
 A: Let's try induction.
The claim is obviously true for $n=1$, so the base case is done.
Now, suppose the claim holds for some $n$.
Let's show it holds for $n+1.$
In order for the claim to fail, we would need
\begin{array}{c}\left(\forall T=\{x_1,x_2,\dots,x_{n+1}\} \subset S\right)\,
\left(\exists \text{ non-negative } a_1,a_2,\dots,a_{n+1} \in \mathbb Q, \text{ not all of them $0$}\right)\,\\
\sum_{i=1}^{n+1}a_ix_i\in \mathbb Q\tag{$*$}\end{array}
Suppose that the statement holds; we will derive a contradiction.
Label $S=\{z_1,z_2,\dots,z_{2n+1}\}$ so that $R=\{z_1,z_2,\dots,z_n\}$ is obtained from our induction hypothesis $($meaning nontrivial non-negative rational linear combinations on $R$ are never rational$)$.
For each $k\in\{1,\dots,n+1\}$ let $T_k=\{z_1,\dots,z_n,z_{n+k}\}$ and write
$$b_kz_{n+k}+\sum_{i=1}^n\,a_{k,i}\hspace{0.2pt}z_i=q_k,\tag{$**$}$$
where $q_k\in \mathbb Q$ and the $a_{k,i}$ and $b_k$ are obtained via $(*)$, meaning they are non-negative rationals and for any given $k$ not all of them vanish.
Notice that our hypothesis on $R$ implies that $b_k\neq 0$.
Finally, let $T^*=\{z_{n+1},z_{n+2},\dots,z_{2n+1}\}$.
By $(*)$, there are non-negative rational $c_i$, not all of them $0$, with $\sum_{i=1}^{n+1}c_iz_{n+i} = q^* \in \mathbb Q$.
Hence
\begin{align}
q^*
&=\sum_{i=1}^{n+1}\frac{c_i}{b_i}\,b_iz_{n+i}\\
&=\sum_{i=1}^{n+1}\frac{c_i}{b_i}\cdot\left(q_i-\sum_{j=1}^n\,a_{i,j}\hspace{0.2pt}z_j\right)
\end{align}
This can be expanded into a nontrivial non-negative rational linear combination on $R$:
\begin{align}
\underbrace{\left(\sum_{i=1}^{n+1}\frac{c_i}{b_i}\,q_i\right)-q^*}_{\in \mathbb Q}
&=\sum_{i=1}^{n+1}\frac{c_i}{b_i}\cdot\sum_{j=1}^n\,a_{i,j}\hspace{0.2pt}z_j\\
&=\sum_{j=1}^{n}\left(\sum_{i=1}^{n+1}\frac{c_i}{b_i}\,a_{i,j}\right) z_j
\end{align}
But the result is rational, which contradicts our choice of $R$. $\square$

EDIT: Here is a short bit showing that not all $\sum_{i=1}^{n+1}\frac{c_i}{b_i}\,a_{i,j}$ vanish.
Notice that for any $j\in\{1,\dots,n\}$ the sum $\sum_{i=1}^{n+1}\frac{c_i}{b_i}\,a_{i,j}$ is over nonnegative rationals, so it is $0$ if and only each term is $0$.
This happens if and only if
$$\forall i\in \{1,\dots,n+1\},\, c_i=0 \lor a_{i,j}=0.$$
Now, not all $c_i$ vanish, so there is some $i_0$ with $c_{i_0}\neq 0$.
In order for the sum $\sum_{i=1}^{n+1}\frac{c_i}{b_i}\,a_{i,j}$ to vanish over all $j$, we would hence need $a_{i_0,j}=0$ for all $j$.
This would imply via $(**)$ that $b_{i_0}z_{n+i_0}=q_{i_0}$, a contradiction.
