= Solution
Contract the two Cartesian indices and use the unit length of each molecular orientation:
$$
\operatorname{Tr}S=\sum_\alpha S_{\alpha\alpha}=\frac1N\sum_i\left(3\sum_{\alpha=x,y,z}s_{i\alpha}^2-\sum_{\alpha=x,y,z}\delta_{\alpha\alpha}\right)=\frac1N\sum_i(3-3)=\boxed{0}.
$$
This is exact for each configuration, with no <ensemble average> assumption. The <nematic order parameter> is also a real <symmetric second-rank tensor>. Under a common <rotation> it transforms as $S\mapsto RSR^{\mathsf T}$, so its powers give <rotational invariants of a symmetric traceless tensor> by the <cyclic property of the trace>. It is unchanged by independent sign flips of the molecular <vectors>, explaining why it is adapted to nematic rather than polar order.
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