If there would exist such a configuration with all distances $>1$ one could place 1981 balls of radius ${1\over2}$ in a cube $Q$ of side length 10 (move all faces of the given cube ${1\over2}$ outwards). But these balls have a total volume of $1037.24\ldots\ $.
In fact the number $1981$ can be improved to $1415$: If there are $N$ points with mutual distance $>1$ in the $9^3$-cube, then $N$ balls of radius ${1\over2}$ fit in the $10^3$-cube $Q$. These balls consume the fraction $\delta={N\cdot{\pi\over6}\over 1000}$ of ${\rm vol}(Q)$. By repeating this configuration periodically one obtains a sphere packing of space with this density; therefore by Hales' theorem one has $\delta\leq{\pi\over 3\sqrt{2}}$ or $N\leq1000\sqrt{2}\doteq1414.2$. This implies that for $N\geq1415$ such a configuration is impossible.