= Solution
The incident <four-momentum> has energy $E$ and spatial momentum $p=\gamma mv$, while the stationary target has energy $mc^2$ and zero spatial momentum. By the <relativistic energy-momentum relation>, $c^2p^2=E^2-m^2c^4$. For the <fixed-target threshold for production of equal-mass particles>, the squared total <centre-of-momentum energy> is
$$
W^2=(E+mc^2)^2-c^2p^2=2mc^2(E+mc^2).
$$
In the <centre-of-momentum frame>, each of the final $N$ particles has energy at least its <rest energy> $mc^2$. <Four-momentum conservation> therefore requires $W\geq Nmc^2$, and thus
$$
\boxed{E\geq\frac{N^2-2}{2}mc^2.}
$$
For the physically relevant $N\geq2$, this is equivalent to $\gamma\geq(N^2-2)/2$. The speed is an increasing function of $\gamma$, so
$$
\frac{v}{c}=\sqrt{1-\gamma^{-2}}
\geq\sqrt{1-\frac4{(N^2-2)^2}}
=\frac{N\sqrt{N^2-4}}{N^2-2}.
$$
Consequently
$$
\boxed{v\geq\frac{N\sqrt{N^2-4}}{N^2-2}\,c.}
$$
Equality is the kinematic threshold at which all final particles are at rest in their common <centre-of-momentum frame>. It shows why a collision with a stationary target needs more incident energy than merely the additional rest energy: some laboratory energy must remain as collective motion. A one-particle final state of the specified mass, with no other outgoing energy, is impossible because its invariant energy would be $mc^2$, whereas the two-particle initial state has invariant energy at least $2mc^2$.
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