Continuing the idea from Gerry's answer. The quantity, you are looking for is just

$$
 D(N) = 2 \left( \\# \{1 \leq n \leq N: \theta n \pmod{1} \in [0,\frac{1}{2}) \}- \frac{N}{2} \right) 
$$
If $\theta = 1/3$, then this quantity grows like $N$. Since $\\#\{ ... \} \sim \frac{2}{3} N$. Something similar happens whenever $\theta = \frac{p}{q}$ with $q$ odd (if I am not mistaken). Of course one cannot achieve a growth of the form $\sim N$ for any irrational number, but one can get arbitrarily close choosing
$$
 \theta = \cfrac{1}{a_1 + \cfrac{1}{a_2 + \dots}}
$$
with the sequence $a_k$ growing fast enough.

In summary, the above strategy should show that given $f(N)$ such that $f(N)/N \to 0$, one can find $\theta$ such that $D(N) \geq c f(N)$ for some small enough $c > 0$ and infinitely many $N$.