The set of reals $a$ which at least one of $a+a_1, a+a_2,\hdots,a+a_n$ is rational is countable, as it's union of $\mathbb{Q}-a_1, \mathbb{Q}-a_2,\hdots,\mathbb{Q}-a_n$. However, since $\mathbb{R}$ is uncountable, there are uncountably infinite choices of $a$.

Suppose there is irrational y so that for k={1,2...n+1}, from the numbers $ky+a_1, ky+a_2...ky+a_n$ there is at least one real. From the box principle for (i;j)={1,2...n+1} there are $iy+a_p$ and $jy+a_p$ that are rational.So (i-j)y is rational i.e. y is rational, contradiction

Suppose that for every real $a$ there exists an $a_i$ for which $a + a_i$ is rational. Now consider the set $S$ of reals that are multiples of some irratoinal number $\alpha$, that is $\{\alpha, 2\alpha, \ldots\}$. For each $k\alpha$ denote $b_k$ as the index for which $k\alpha + a_{b_k}$ is rational. Since $S$ is infinite there exists $k_1\alpha$ and $k_2\alpha$ with $k_1 \neq k_2$ for which they have the same index $b_{k_1} = b_{k_2}$. We have that\[k_1\alpha + g = \text{rational}\]\[k_2\alpha + g = \text{rational}\]where $g = a_{b_{k_1}} = a_{b_{k_2}}$. Upon subtracting we see that $(k_1 - k_2)\alpha$ is rational which is possible only if $\alpha$ is rational, a clear contradiction. Hence our original assumption was false and in $S$ there exists a $k\alpha = a$ that makes the statement true. $\blacksquare$