For a positive integer $n,$ simplify
\[1^2 - 2^2 + 3^2 - 4^2 + \dots + (2n - 1)^2 - (2n)^2.\]
We can pair the terms and use the difference of squares factorization, to get
\begin{align*}
&(1^2 - 2^2) + (3^2 - 4^2) + \dots + [(2n - 1)^2 - (2n)^2] \\
&= (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + \dots + [(2n - 1) - (2n)][(2n - 1) + (2n)] \\
&= (-1)(1 + 2) + (-1)(3 + 4) + \dots + (-1)[(2n - 1) + (2n)] \\
&= -1 - 2 - 3 - 4 - \dots - (2n - 1) - 2n \\
&= -\frac{2n(2n + 1)}{2} \\
&= \boxed{-2n^2 - n}.
\end{align*}