$\newcommand{mod}{\mathrm{mod} \ }$
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
$$f(t) := \min_i f_i(t)$$

$\hskip2in$

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). For example, let
$f_i(t_0)= m_it_0-k_i$ and $f_j(t_0)=-m_j t_0 +(k_j+1)$, for some $k_i,k_j \geq 0$. Now the equality $f_i(t_0)=f_j(t_0)$ gives
$$t_0 = \frac{k_i+k_j + 1}{m_i+m_j},\quad \kappa=f(t_0)= \frac{m_ik_j-k_im_j+m_i}{m_i+m_j}$$
Now, the result at hand.

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