Recall that a compactification of a topological space $X$ is a pair $(K,h)$ such that $K$ is a compact space and $h:X\to K$ is an embedding such that $h(X)$ is a dense subspace of $K$. The remainder is the set $K\setminus h(X)$.
Is there a Hausdorff compactification of the irrational numbers with its usual topology such that the remainder is a discrete space?
No such compactification exists. More generally, suppose $X$ is any space in which no point has a compact neighborhood and $K$ is a Hausdorff compactification of $X$ (without loss of generality taking $X\subset K$ and $h$ the inclusion map). Suppose $k\in K\setminus X$ is an isolated point. Then there is some open set $U\subseteq K$ containing $k$ and no other points of $K\setminus X$. Since $X$ is dense in $K$, $U$ contains some point $x\in X$. Since no neighborhood of $x$ in $X$ is compact, every neighborhood of $x$ in $K$ must intersect $K\setminus X$. In particular, $(U\setminus \{k\})\cap(K\setminus X)\neq\emptyset$. This contradicts our choice of $U$. Thus $K\setminus X$ cannot have any isolated points.