Here is a probabilistic way. Consider an i.i.d. sequence $(X_n)_{n\geqslant1}$ of Bernoulli random variables such that $\mathbb P(X_n=+1)=\mathbb P(X_n=-1)=\frac12$. Then $X=\sum\limits_{n\geqslant1}2^{-n}X_n$ is uniformly distributed on $[-1,1]$ hence $X$ is irrational with probability $1$. And naturally, $\sum\limits_{n\geqslant1}2^{-n}|X_n|=1$.
A deterministic way is as follows. Pick any irrational number $z$ in $(0,1)$, with binary expansion $z=\sum\limits_{n\geqslant1}z_n2^{-n}$, where each $z_n$ is in $\{0,1\}$. For every $n\geqslant1$, consider $x_n=(2z_n-1)2^{-n}$. Then $\sum\limits_{n\geqslant1}x_n=2z-1$ is irrational and $|2z_n-1|=1$ for every $n\geqslant1$ hence $\sum\limits_{n\geqslant1}|x_n|=1$.