Let us examine each complex using Valence Bond Theory:

1. $[Ni(NH_3)_6]^{2+}$: The central metal ion is $Ni^{2+}$ with a $3d^8$ configuration. To form an octahedral complex, the two $4d$ orbitals are used for hybridization, resulting in $sp^3d^2$ (outer orbital complex). It has two unpaired electrons in the $3d$ level, making it paramagnetic.
2. $[Zn(NH_3)_6]^{2+}$: The central metal ion is $Zn^{2+}$ with a $3d^{10}$ configuration. It uses outer $4d$ orbitals for $sp^3d^2$ hybridization (outer orbital complex), but all electrons are paired, making it diamagnetic.
3. $[Cr(NH_3)_6]^{3+}$: The central metal ion is $Cr^{3+}$ with a $3d^3$ configuration. The two inner $3d$ orbitals are available for $d^2sp^3$ hybridization (inner orbital complex). It has three unpaired electrons, making it paramagnetic.
4. $[Co(NH_3)_6]^{3+}$: The central metal ion is $Co^{3+}$ with a $3d^6$ configuration. Ammonia ($NH_3$) acts as a strong field ligand, forcing electrons to pair up. This makes two inner $3d$ orbitals available for $d^2sp^3$ hybridization (inner orbital complex) and results in zero unpaired electrons, making it diamagnetic.

Therefore, $[Ni(NH_3)_6]^{2+}$ is the only complex that is both an outer orbital complex and paramagnetic.