Consider $y_1 < y_2 < \dots < y_n < y_{n+1}$, irrational numbers such that $y_j - y_i$ is irrational for all $1 \le i < j \le n + 1$. (One could take, for example,

$y < 2y < 3y < \dots < (n+1)y$, where $y$ is irrational.) We plan to prove that one of these irrational numbers can be chosen as $x$.

Assume the contrary to be true, i.e. for each $y_k$, at least one of the numbers $y_k + a_1, y_k + a_2, \dots, y_k + a_n$ is rational. But there are $n+1$ choices for $y_k$, and only $n$ choices for $a_m$ such that $y_k + a_m$ is rational. By the Pigeon Principle, it follows that there must be an index $m$ and two irrational numbers $y_i, y_j$ such that $a_m + y_i$ and $a_m + y_j$ are both rational. But this would mean that their difference, $y_j - y_i$, is also rational, which contradicts the choice of the numbers $y_k$.