If $$gcd(m, n) = 1$$ and $$1^2 - 2^2 + 3^2 - 4^2 + \ldots + (2021)^2 - (2022)^2 + (2023)^2 = 1012m^2n$$ then $$m^2 - n^2$$ is equal to

We need to evaluate $$1^2 - 2^2 + 3^2 - 4^2 + \cdots + (2021)^2 - (2022)^2 + (2023)^2$$.

Pairing consecutive terms and using the identity $$k^2 - (k+1)^2 = -(2k+1)$$, we write the sum as:

Each pair yields $$k^2 - (k+1)^2 = -(2k+1)$$ for $$k = 1, 3, 5, \ldots, 2021$$, so the values are $$-3, -7, -11, \ldots, -4043$$ and there are $$1011$$ such pairs.

Hence the sum of these pairs is $$-(3 + 7 + 11 + \cdots + 4043)$$, which is an arithmetic progression with first term 3, last term 4043, and 1011 terms, giving

Adding the final term, we get

Since $$S = 1012 \cdot m^2 \cdot n$$, it follows that $$m^2 n = 2023$$. Factoring 2023 gives $$2023 = 7 \times 17^2$$, and with $$\gcd(m, n) = 1$$ we have $$m = 17$$ and $$n = 7$$.

Therefore,

The correct answer is Option A: 240.