Let $N_b = n^2$ and $k = (N_b - 1)/2$. There are $\binom{N_b}{2}$ ways of placing 2 identical red checkers, and for any such placement there are $\binom{N_b-2}{2}$ ways of next placing the 2 identical blue checkers. Therefore the total number of ways of placing 2 identical red checkers and 2 identical blue checkers on the board is
$$N_t = \binom{N_b}{2} \binom{N_b-2}{2} = \frac{N_b(N_b-1)(N_b-2)(N_b-3)}{4} \\ = N_b \cdot (N_b-2) \cdot k \cdot (k-1).$$
Some of these checker placements have $180^\circ$ symmetry. The number of such placements, denoted $N_s$, is equal to the number of ways of placing a red checker and a blue checker on the "semi-board" formed by taking any subset of $k = (N_b - 1)/2$ squares such that the centre square is omitted, and no square in the subset is the image of another square under central symmetry in the central square (to complete the symmetrical placement, the missing red checker is placed on the image square of the placed red checker, and similarly for the missing blue checker). It follows that the number of $180^\circ-symmetrical$ checker placements is
$$N_s = k(k-1).$$
There are $N_t - N_s$ placements which do not possess $180^\circ$ symmetry; such placements are counted four times, while the $N_s$ $180^\circ-symmetrical$ placements are counted twice. Thus
$$\begin{aligned} N &= \frac{N_t - N_s}{4} + \frac{N_s}{2} = \frac{N_t + N_s}{4} \\ &= k(k-1)\frac{N_b(N_b-2)+1}{4} = k(k-1) \cdot k^2 = k^4 - k^3, \end{aligned}$$
and the result is proved.