nodal analysis, current direction
Indeed you can define the current direction in which ever way you want. If you want to get the current going into the node N2, it should be \$\frac{0-v_2}{12k} = -\frac{v_2}{12k}\$.
If you add this current, then you get the same equation:
$$\frac{(v1-v2)}{6000}\ - 4mA + (\frac{-v2}{12000})\ = 0$$