The condition says adjacent cells have different residues mod $3.$ Each residue class among $1, \ldots, 12$ has exactly $4$ members, so $N = K \cdot (4!)^3,$ where $K$ is the number of ways to fill the grid with residues $0, 1, 2,$ each used $4$ times, with adjacent cells different.

A column is an ordered pair $(a, b)$ of distinct residues. If the current column is $(a, b)$ and $e$ is the third residue, the next column must be one of $(b, a),$ $(b, e),$ $(e, a):$ each of the three unordered pairs $\{a,b\},$ $\{b,e\},$ $\{a,e\}$ occurs in exactly one allowed orientation. So a residue pattern is determined by the sequence of six unordered pairs together with the orientation of the first column. Since each residue must appear $4$ times, each of the three pairs must be used exactly twice, giving $\frac{6!}{2!\,2!\,2!} = 90$ sequences and $K = 2 \cdot 90 = 180.$

Therefore $N = 180 \cdot 24^3$ $= 2{,}488{,}320$ $= 2^{11} \cdot 3^5 \cdot 5,$ which has $12 \cdot 6 \cdot 2 = 144$ positive divisors.