1. We need to find $S=\{n\in \mathbb Z: |f(n)|\le 800\}, \quad f(x)=2x^2-x-1.$ Then compute $\sum_{n\in S} f(n).$

2. First solve the inequality $|2n^2-n-1|\le 800.$ This means $-800\le 2n^2-n-1\le 800.$ So we solve both parts.

(i) Upper bound

$2n^2-n-1\le 800$ $2n^2-n-801\le 0.$ Solve the quadratic equation: $2n^2-n-801=0.$ Discriminant: $D=1+4\cdot 2\cdot 801=1+6408=6409=81^2.$ Hence, $n=\frac{1\pm 81}{4}.$ So the roots are $n=-20,\quad n=\frac{41}{2}=20.5.$ Since the parabola opens upward, $2n^2-n-801\le 0 \iff -20\le n\le 20.5.$ For integer $n$, $-20\le n\le 20.$

(ii) Lower bound

$2n^2-n-1\ge -800$ $2n^2-n+799\ge 0.$ Now check its discriminant: $D=(-1)^2-4\cdot 2\cdot 799=1-6392=-6391<0.$ Since the leading coefficient is positive and discriminant is negative, $2n^2-n+799>0$ for all real $n$. So this inequality is always true.

Therefore, $S=\{-20,-19,\dots,20\}.$

3. Now compute $\sum_{n=-20}^{20}(2n^2-n-1).$ Split the sum: $\sum_{n=-20}^{20}(2n^2-n-1)=2\sum_{n=-20}^{20}n^2-\sum_{n=-20}^{20}n-\sum_{n=-20}^{20}1.$

Using symmetry, $\sum_{n=-20}^{20} n=0.$ Also there are $20-(-20)+1=41$ integers, so $\sum_{n=-20}^{20}1=41.$

Now, $\sum_{n=-20}^{20} n^2=2\sum_{n=1}^{20} n^2.$ And $\sum_{n=1}^{20} n^2=\frac{20\cdot 21\cdot 41}{6}=2870.$ Hence, $\sum_{n=-20}^{20} n^2=2\cdot 2870=5740.$

Therefore, $\sum_{n=-20}^{20}(2n^2-n-1)=2(5740)-0-41=11480-41=11439.$

4. Final answer: $\boxed{11439}$

5. Comparison with stored answer: Stored correct answer is $10620$, but the correct computed value is $11439$. So I disagree with the stored answer.