 Solution

Orient the $x$ axis along the displacement from the earlier event to the later one, so $\Delta x=d=1000\mathrm{m}$ and $\Delta t=10^{-6}\mathrm{s}$. For a parallel Lorentz transformation, simultaneity in the second inertial frame requiresThe second inertial frame moves towards the later event's location. The positive value $d^2-c^2\Delta t^2$ establishes spacelike separation, which permits this change of simultaneity. For a general velocity $\mathbf{V}$, the condition is $\mathbf{V}\cdot \Delta \mathbf{r}=c^2\Delta t$: only its component along the separation is fixed at the displayed value, and transverse components are allowed provided $|\mathbf{V}|<c$. Thus the parallel velocity has the smallest possible speed.

A physical inertial frame must have $|v|<c$, givingThere is no attained maximum: $d/c$ is the supremum. For the concrete boundary case $\Delta t=1000\mathrm{m}/c$, the events have null separation and the formula would require $v=c$; no inertial frame can make them simultaneous.