Let $a_1$ be an irrational number in $A$. If $A$ contains infinitely many irrational numbers, then for any $n$ we can pick $n$ distinct irrational numbers from $A$; their sum is irrational (since the sum of irrational numbers is irrational unless they sum to a rational, but with infinitely many choices, we can avoid this).

Suppose $A$ contains only finitely many irrational numbers. Then $A$ contains infinitely many rational numbers. Let $a_1$ be an irrational number in $A$, and $a_2, \ldots, a_n$ be any $n-1$ rational numbers in $A$. Then $a_1 + a_2 + \cdots + a_n$ is irrational (since the sum of an irrational and any number of rationals is irrational).

Therefore, for any positive integer $n$, we can find $n$ elements of $A$ whose sum is irrational.