Not even Wolfram-Alpha is going to help!

A solution without using complex numbers. First, I will prove the following two lemmas.

Lemma 1: If $\sin x$ is a number distinct from $0$ then $\: \sum_{k=1}^{n} \cos(k x)=\frac{ \sin(n+1)x+\sin nx -sin x}{2\sin x}$

Proof: $\sum_{k=1}^{n} \cos(k x)=\frac{1}{2\sin x} \sum_{k=1}^{n} 2\sin x\cos(k x)$ $=\frac{1}{2\sin x}\sum_{k=1}^{n}(\sin (k+1)x - \sin (k-1)x )$ $=\frac{1}{2\sin x}[(\sin (n+1)x +\sin nx )-(\sin x +\sin 0 )$ ] $=\frac{ \sin(n+1)x+\sin nx -sin x}{2\sin x. }$ Q. E.D.

Lemma 2 : If $x=\frac{2\pi j}{2n+1}$ where $j$ is any integer, $n$ is a natural number and $\sin x$ is different from $0$ , then $\sum_{k=1}^{n} \cos(k x)=-\frac{1}{2}$

Proof: For such a value of $x$ it is always true that $\sin (n+1)x +\sin nx =0$ . Indeed, $\sin (n+1)x +\sin nx =2 \sin (\frac{(n+1)x+nx}{2})\cos(\frac{(n+1)x-nx}{2})=$ $=2 \sin \pi j\cos(\frac{(n+1)x-nx}{2})=0.$

So we can complete the proof of Lemma 2 using the Lemma 1.

Q. E. D

Now using Lemma 2 twice we have that

$\sum_{1\le k<s\le n} \cos \frac{2\pi k}{2n+1}\cos \frac{2\pi s}{2n+1}=\frac{1}{2}((\sum_{k=1}^{n} \cos\frac{2\pi k}{2n+1})^{2}-\sum_{k=1}^{n} \cos^{2}\frac{2\pi k}{2n+1})$ $=\frac{1}{2} (\frac{1}{4}-\sum_{k=1}^{n} \frac{1+\cos\frac{4\pi k}{2n+1}}{2})=\frac{1}{2} (\frac{1}{4}-\frac{n}{2}-\frac{1}{2}\sum_{k=1}^{n} \cos\frac{4\pi k}{2n+1})$ $=\frac{1}{2} (\frac{1}{4}-\frac{n}{2}-\frac{1}{2}(-\frac{1}{2}))=\frac{1-n}{4}.$

Therefore the answer of our question can be found by making $n=1007$ in the latter expression and multiplying the result by ten. So we get $10 (\frac{1-1007}{4})=-2515.$

Actually, it is $4*\sum_{1\le k<s\le n}\cos(\frac{2\pi k }{2n+1})\cos(\frac{2\pi s }{2n+1})+n$ . You can understand it by finding the coefficient of $x^2$ of a polynomial of the same type with less number of factors. For example, what is the coefficient of $x^2$ for the polynomial $(x^2-2p_{1}x+1)(x^2-2p_{2}x+1)(x^2-2p_{3}x+1)$ ? The term containing $x^2$ is obtained when you do the following:

( $x^2$ from the first factor)X ( 1 in the second) X( 1 in the third) + (1 from the first )x( $x^2$ from the second)x(1 from the third)+(1 from the first )x(1 from the second)x( $x^2$ from the third) +( $-2p_{1}x$ from the first) x( $-2p_{2}x$ from the second)x(1 from the third)+( $-2p_{1}x$ from the first) x(1 from the second)x( $-2p_{3}x$ from the third)+(1 from the first) x( $-2p_{2}x$ from the second)x( $-2p_{3}x$ from the third) .

In this case you would obtain: $(1+1+1+4 p_{1}p_{2}+4 p_{1}p_{3}+4 p_{2}p_{3})x^2$ $=(3+4 p_{1}p_{2}+4 p_{1}p_{3}+4 p_{2}p_{3})x^2.$ Then the coefficient of $x^2$ would be $3+4 p_{1}p_{2}+4 p_{1}p_{3}+4 p_{2}p_{3}.$