= Solution
Use $c=1$ and define the total <four-momentum> $P^\mu=(E,\mathbf P)=\sum_i(E_i,\mathbf p_i)$. For an <inertial frame> moving with <velocity> $\mathbf V$ relative to the laboratory, the <Lorentz boost> gives
$$
P'_{\parallel}=\gamma_V(P_{\parallel}-VE),\qquad \mathbf P'_{\perp}=\mathbf P_{\perp}.
$$
Choose $\mathbf V$ parallel to $\mathbf P$ and require zero total <three-momentum>. Then
$$
\boxed{\mathbf V_{\mathrm{CM/lab}}=\frac{\mathbf P}{E}=\frac{\sum_i\mathbf p_i}{\sum_iE_i}.}
$$
This requires $E>|\mathbf P|$, so that the total <four-momentum> is timelike. Its <invariant mass> is $\sqrt{E^2-|\mathbf P|^2}$, the total <relativistic energy> in the <center-of-momentum frame>. If all particles are exactly parallel <photons>, the total <four-momentum> is null and no finite-<velocity> <center-of-momentum frame> exists.
\b[Sign qualification.] The minus sign printed in the source describes the laboratory's <velocity> in the <center-of-momentum frame>, or the parameter of an active <Lorentz boost> with a plus-sign spatial rule. It is not the <velocity> of the <center-of-momentum frame> relative to the laboratory under question 1's passive convention. For example, a single particle with $(E,P_x)=(5,3)$ has its rest frame moving at $V_x=3/5$. The printed $V_x=-3/5$ would give $P'_x=(5/4)(3+3)=15/2$, which is not zero. This explicitly resolves the sign rather than assigning a rest frame moving away from the particle.
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