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

Let

$S=1^2-2^2+3^2-4^2+....+{\left(2021\right)}^2-{\left(2022\right)}^2+{\left(2023\right)}^2$

$\Rightarrow S=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+....+\left(2021-2022\right)\left(2021+2022\right)+{\left(2023\right)}^2$

$\Rightarrow S=-\left[3+7+11+15+....+4043\right]+{\left(2023\right)}^2$

The number of terms in the bracket are $\frac{2022}{2}=1011$$\Rightarrow S=-\frac{1011}{2}\left(6+1010\times 4\right)+{\left(2023\right)}^2$

$\Rightarrow S=-1011\times 2023+{\left(2023\right)}^2$

$\Rightarrow S=2023\times 1012$

$\Rightarrow S={17}^2\times 7\times 1012$

So,

$m=17, n=7$ and $gcd\left(17,7\right)=1$

Hence, $m^2-n^2={17}^2-7^2=240$

Hence this is the correct option.