To find the maximum number of isomers, we evaluate each given complex:

1. $[Co(NH_3)_4Cl_2]$: It is an octahedral complex of the type $[Ma_4b_2]$ and exhibits geometrical isomerism, existing as cis and trans isomers. Total isomers = 2.
2. $[Ni(en)(NH_3)_4]^{2+}$: The bidentate ligand 'en' can only occupy adjacent (cis) positions. Since all 4 remaining positions are occupied by identical $NH_3$ ligands, it does not show geometrical or optical isomerism. Total isomers = 1.
3. $[Ni(C_2O_4)(en)_2]$: It is an octahedral complex with three bidentate ligands, similar to the $[M(AA)_3]$ type. It lacks a plane of symmetry and exists as a pair of optical isomers (non-superimposable $d$ and $l$ enantiomers). Total isomers = 2.
4. $[Cr(SCN)_2(NH_3)_4]^+$: It is of the type $[Ma_4b_2]$ and shows cis and trans geometrical isomerism. Additionally, the $SCN^-$ ion is an ambidentate ligand, meaning it can bind through sulphur (–SCN) or nitrogen (–NCS), leading to linkage isomerism. The possible combinations of linkage isomers are: (i) both bound via S (SCN, SCN), (ii) both bound via N (NCS, NCS), and (iii) one S-bound and one N-bound (SCN, NCS). Each of these 3 combinations can exist as cis and trans forms. Total isomers = $3 \times 2 = 6$.

Therefore, $[Cr(SCN)_2(NH_3)_4]^+$ gives the maximum number of isomers.