Calculate the sum of the coordinates of all pairs of positive integers $(n, k)$ such that $k\equiv 0, 3\pmod 4$, $n > k$, and $\displaystyle\sum^n_{i = k + 1} i^3 = (96^2\cdot3 - 1)\displaystyle\left(\sum^k_{i = 1} i\right)^2 + 48^2$

Let's think step by step and output the final answer within \boxed{}.
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