Define $f_i(t) := \| t m_i \|$. We have
$$ f_i(t) = \begin{cases} m_i t - k, & \text{if } t\in[\frac{2k}{2m_i},\frac{2k+1}{2m_i})\\ -m_i t +(k+1), & \text{if } t \in [\frac{2k+1}{2m_i},\frac{2k+2}{2m_i}) \end{cases} \tag{1} $$ $f_i$ is a continuous piecewise linear function, which is $0$ at $\frac{2k}{2m_i}$ and is $\frac 12$ at $\frac{2k+1}{2m_i}$.
Now, consider the function $$f(t) := \min_i f_i(t)$$ $\newcommand{mod}{\mathrm{mod} \ }$ (Considering the graph of $f$ is helpful). We want to find $\sup_{t >0} f(t)$. Since $f$ is periodic, it is enough to consider $\sup_{0<t<1} f(t)$. Moreover, for sufficiently small $\epsilon >0$, it is enough to consider $\sup_{\epsilon \leq t \leq1-\epsilon} f(t)$. Since $f$ is a continuous function attains its supremum on $\epsilon \leq t \leq1-\epsilon$ at some point, like $t_0$. There is two function, like $f_i$ and $f_j$, such that $f_i(t_0)=f_j(t_0)=f(t_0)$. Note that the rules of $f_i$ and $f_j$ at $t_0$ are different in (1), i.e. $\lfloor 2 m_it_0 \rfloor \not\equiv \lfloor 2 m_j t_0 \rfloor (\mod 2)$. (otherwise $f(t_0)$ can not be supremum).
Now, the result at hand.