6.6 Calculate

$$
\begin{aligned}
& \left(1+3^{2}+5^{2}+\ldots+(2 n-1)^{2}+\ldots+199^{2}\right)- \\
& -\left(2^{2}+4^{2}+6^{2}+\ldots+(2 n)^{2}+\ldots+200^{2}\right)
\end{aligned}
$$
6.6 Note that $(2 n)^{2}-(2 n-1)^{2}=4 n-1$. The number $n$ can be found from the equation $4 n-1=200^{2}-199^{2}$, or $4 n-1=399$, from which $n=100$. We will show that the sequence $a_{n}=(2 n)^{2}-(2 n-1)^{2}$ is an arithmetic progression. Indeed,

$a_{n}-a_{n-1}=\left((2 n)^{2}-(2 n-1)^{2}\right)-\left((2 n-2)^{2}-(2 n-3)^{2}\right)=$

$=4 n-1-4 n+5=4$,

i.e., $d=4$. Now we find

$-S_{100}=\frac{3+399}{3} \cdot 100=20100$, from which $S=-20100$.

Answer: -20100.