Solution:

Let the given numbers be $a_{1}, a_{2}, \ldots, a_{2n+1}$. Choose first 1 and then choose at any step (if it is possible) a number that is not a linear combination with rational coefficients of the already chosen numbers. We may assume that the chosen numbers are $a_{0}=1, a_{1}, a_{2}, \ldots, a_{k}$, $1 \leq k \leq 2n+1$. It is easy to see that any linear combination with rational coefficients of the given numbers can be uniquely presented as a linear combination of these numbers.

Let $a_{i}=\sum_{j=0}^{k} \alpha_{ij} a_{j}$, where $\alpha_{ij} \in \mathbb{Q}$, $1 \leq i \leq 2n+1$, $0 \leq j \leq k$. Then a sum of $a_{i}$'s is a rational number if and only if the sum of the corresponding numbers $b_{i}=a_{i}-\alpha_{i0}$ vanishes. Since $b_{1}, b_{2}, \ldots, b_{2n+1}$ are irrational numbers, they are non-zero. In particular, at least $n+1$ of them have the same sign and hence the corresponding $a_{i}$'s have the desired property.