Rod is dead!

After collision let the velocity of centre of mass be $v$ and angular velocity about the centre of the rod be $\omega$

Apply angular momentum conservation about the centre of the rod to get

$mv_0x = \frac{ML^2 \omega}{12}$

$\implies \dfrac{ \color{#D61F06}{ \omega L} }{2} = \dfrac{ 6 mv_0 x } {ML} = \dfrac{6mv_0 }{Mn}$ ........... . $\color{#3D99F6}{ ( \text{using} ~L = nx ) }$

conserve momentum to get

$mv_0 = Mv \implies \color{#D61F06}{v} = \frac{m v_0 }{M}$

The velocity of the point at the farther end of the rod is in the direction opposite to $v_0$

$\displaystyle \therefore \frac{ \omega L }{2} - v > 0$

$\displaystyle \implies \frac{6 mv_0 }{Mn} - \frac{mv_0 }{M} > 0$

$\displaystyle \frac{6mv_0 }{Mn} > \frac{mv_0 }{M}$

this gives $\color{#3D99F6}{ n < 6}$

In this scenario,angular momentum is conserved about the centre of the rod.So ,

$mv_{o}x$ = $Iω$

$mv_{o}x$ = $\frac{ML^{2}ω}{12}$

$mv_{o}x$ = $\frac{2ML^{2}v_{o}}{12L}$

$mv_{o}x$ = $\frac{MLv_{o}}{6}$

$mv_{o}x$ = $\frac{Mnxv_{o}}{6}$

$m$ = $\frac{Mn}{6}$

$n$ = $\frac{6m}{M}$

As M > m,

$\frac{m}{M}$ < $1$

$\frac{6m}{M}$ < $6$

$\boxed{n < 6}$