= Solution
With the reversed <electric field>, reflection about $x=d/2$ exchanges the <electron> and <Positron>. They therefore meet there with equal and opposite <momenta>. Each travels $d/2$, and the <work-energy theorem> gives its <kinetic energy>:
$$
K=eE\frac d2=e(5.0\times10^6\,\mathrm{V\,m^{-1}})(0.100\,\mathrm m),\qquad
\boxed{K_e=K_p=0.500\,\mathrm{MeV}.}
$$
Their combined <kinetic energy> is $1.000\,\mathrm{MeV}$, while each total <relativistic energy> is $\mathcal E=0.511+0.500=1.011\,\mathrm{MeV}$. The laboratory is the <center-of-momentum frame> of the colliding pair.
Local <four-momentum conservation> in the <electron-positron annihilation> excludes one real <photon>: the incoming total <four-momentum> is timelike, with zero spatial component and positive <energy>, while a single <photon>'s <four-momentum> is null and has $|\mathbf p|=\mathcal E/c>0$. Two <photons> can balance their <momenta> by travelling in opposite directions, so
$$
\boxed{N_{\min}=2,\qquad \mathcal E_\gamma=1.011\,\mathrm{MeV},\qquad
\lambda=\frac{hc}{\mathcal E_\gamma}\simeq1.226\times10^{-12}\,\mathrm m.}
$$
The equal <photon> <energies> follow from opposite equal <momenta>, not merely from sharing the incoming <energy> arbitrarily. Here $hc\simeq1.23984\,\mathrm{MeV\,pm}$. This is the free local annihilation idealization implicit in the question; significant <momentum> transfer to the external apparatus would change the kinematic system being considered.
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