New kind of series!

The sum given can be expressed as $S=\displaystyle \sum_{n=1}^{100} (-1)^n s_n$ , where $s_n$ is the sum of integers lesser than or equal to perfect square $n^2$ but greater than $(n-1)^2$ . Therefore,

$\begin{aligned} S & = \sum_{n=1}^{100} (-1)\color{#3D99F6} s_n \\ & = \sum_{n=1}^{100} (-1)^n\color{#3D99F6} \sum_{k=1}^{n^2-(n-1)^2} \left((n-1)^2+k\right) \\ & = \sum_{n=1}^{100} (-1)^n \left((n-1)^2 \left(n^2-(n-1)^2\right)+\frac 12\left(n^2-(n-1)^2\right)\left(n^2-(n-1)^2+1\right)\right) \\ & = \sum_{n=1}^{100} (-1)^n \left(\left(n^2-2n+1\right) (2n-1)+\frac 12(2n-1)(2n)\right) \\ & = \sum_{n=1}^{100} (-1)^n (2n-1)\left(n^2-n+1\right) \\ & = \sum_{n=1}^{100} (-1)^n \left(2n^2-3n^2+3n-1\right) \\ & = \sum_{n=1}^{100} (-1)^n \left((n-1)^3+n^3\right) \\ & = - 0^3-\cancel{1^3}+\cancel{1^3}+\cancel{2^3}-\cancel{2^3}-\cancel{3^3}+\cdots+\cancel{99^3} + 100^3 \\ & = \boxed{1000000} \end{aligned}$