= Solution
With $c=1$, the total <four-momentum> is $P_{\mathrm{tot}}=\sum_iP_i$. A sum of <Lorentz four-vectors> transforms as a four-vector, so
$$
\boxed{s^2=P_{\mathrm{tot}}\cdot P_{\mathrm{tot}}
=\left(\sum_iW_i\right)^2-\left|\sum_i\mathbf p_i\right|^2}
$$
is a <Lorentz invariant>. Its value is the squared <invariant mass> of the complete system, irrespective of whether its constituents interact. Isolation makes the total <four-momentum> constant in time.
For a <photon gas>, write $\mathbf p_i=W_i\mathbf n_i$ with $|\mathbf n_i|=1$. Its individual null norms do not make the total norm vanish:
$$
\boxed{s^2=2\sum_{i<j}W_iW_j(1-\mathbf n_i\cdot\mathbf n_j)\geq0.}
$$
For an isotropic gas in its <momentum> rest frame, $\sum_i\mathbf p_i=0$ and $s^2=W_{\mathrm{tot}}^2$. Zero is possible only when all nonzero <photons> propagate in the same direction. Two equal-energy <photons> at right angles, for example, give $s^2=2W^2$, rather than zero. Without an angular distribution, the pairwise expression is the general answer; isotropy is an additional assumption.
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