= Solution
The equal and opposite incoming <momenta> make the laboratory the <centre-of-mass frame>. The initial total <four-momentum> is timelike. One outgoing <photon> would have null <four-momentum>, so an isolated two-body annihilation cannot produce exactly one <photon>. Two <photons> are kinematically possible and must have opposite <momenta> and equal <energies>:
$$
\boxed{N_{\gamma,\min}=2,\qquad
E_\gamma=\frac{K_{\mathrm{tot}}+2m_ec^2}{2}
=500.511\,\mathrm{MeV}.}
$$
The quoted $0.511\,\mathrm{MeV}$ is the rest <energy> $m_ec^2$, not a mass in SI units. The <energy> balance includes both rest <energies> as well as the $1\,\mathrm{GeV}$ total <kinetic energy>.
This minimum concerns the complete final state. A detector with incomplete acceptance could register only one of the two <photons>, or none; conservation laws do not impose a minimum detected count without an ideal-detection assumption. Additional outgoing <photons> are also possible.
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