In general: no.
If $a<0$, consider $\epsilon=3\pi|a|$. Then for all $n\in\mathbb N$, $b\in\mathbb N$, we have $|b-a(2n+1)\pi|\ge 3\pi |a|=\epsilon$.
Otherwise (e.g. if you switch to $\mathbb Z$ instead of $\mathbb N$), the irrationality of $a$ does not play a role. The answer depends on whether or note $\pi a$ is irrational. If $a\pi$ is rational, we may have bad luck (namely when the denominator in shortest terms is even).
Assume therefore that $a\pi$ is irrational.
By the density of $\{\,ka\pi\bmod 1\mid k\in\mathbb Z\,\}$ in $[0,1]$ we can surely find $b$ and $k$ such that $|b-ak\pi|<\epsilon$. However, can we ensure that $k$ is odd?
Yes: Find $k$ with $0<ak\pi\bmod 1<\epsilon$. If $k$ is odd, we are done. Otherwise, the numbers $rak\pi\bmod 1$, $r\in \mathbb N$, walk trhogh $[0,1]$ in steps of size $<\epsilon$, hence for suitable $r$ differ from $a\pi\bmod 1$ by less than $\epsilon$. We conclude that $(rk-1)a\pi$ differes from an integer by less than $\epsilon$.