jmc

Let $23a$ be the multiple we seek. Thus $$23a\equiv 4(\mathrm{mod}89).$$ Multiplying both sides of this equation by $4$, then reducing modulo $89$, gives: $$\begin{array}{rl}92a& \equiv 16(\mathrm{mod}89)\\ 3a& \equiv 16(\mathrm{mod}89)\end{array}$$ Multiplying both sides by $30$, then reducing again, gives: $$\begin{array}{rl}90a& \equiv 480(\mathrm{mod}89)\\ a& \equiv 480-445=35(\mathrm{mod}89)\end{array}$$ All of these steps are reversible, so $35$ is the unique solution $(\mathrm{mod}89)$ to the original congruence. The smallest positive solution is $a=35$, giving $23a=\boxed{805}$. (Indeed, we can check that $805=9\cdot 89+4$.)