, 2021

Solution:
For any positive integers $a$ and $b$, the sum of the first $a$ multiples of $b$ is $b + 2b + \cdots + ab = b(1 + 2 + \cdots + a) = \frac{a(a+1)b}{2}$. Thus, the conditions imply $m(m+1)n = 240$ and $m^{3}(m^{3}+1)n^{3} = 8064000$, whence
$$\frac{(m+1)^{3}}{m^{3}+1} = \frac{(m(m+1)n)^{3}}{m^{3}(m^{3}+1)n^{3}} = \frac{240^{3}}{8064000} = \frac{12}{7}$$
Thus, we have $7(m+1)^{2} = 12(m^{2} - m + 1)$ or $5m^{2} - 26m + 5 = 0$, so $m = 5$ and therefore $n = 8$. The answer is $\frac{m^{2}(m^{2}+1)}{2} n^{2} = 20800$.