As if they were all alike

Denoting $j$ as the imaginary unit, we have:

$\large = \sum_{n=0}^{999} \Big( \frac{e^{jn} + e^{-jn}}{2} \Big) ^2$

$\large = \sum_{n=0}^{999} \frac{e^{j2n} + e^{-j2n} + 2}{4}$

$\large = \frac14 (\sum_{n=0}^{999} e^{j2n} + \sum_{n=0}^{999} e^{-j2n} + \sum_{n=0}^{999} 2 )$

$\large = \frac14 ( \frac{e^{j2000} - 1}{e^{j2}-1} + \frac{e^{-j2000} - 1}{e^{-j2}-1} + 2000 )$

$\large = \frac14 ( \frac{e^{j1998} + e^{-j1998} - e^{j2} - e^{-j2} - e^{j2000} - e^{-j2000} + 2}{2 - e^{j2} - e^{-j2}} ) + 500$

$\large = \frac14 (\frac{cos(1998) - cos(2000) - cos(2) + 1}{1 - cos(2)}) + 500$

$\large = \fbox{500.4912}$

So:

$\large cos(\frac{\pi}{\alpha})^2 = 0.5004912$

$\large cos(\frac{\pi}{\alpha}) = 0.7074540$

$\frac{\pi}{\alpha} = 0.7849070$

$\color{#3D99F6} \alpha = \fbox{4.0025030}$

Making our answer $\color{#3D99F6} \fbox{4}$

NOTE: This isn't a formal solution (see Guilherme's solution for a formal one), but it's an interesting heuristic.

Since $\cos^2(x)$ has fundamental period $\pi$ , we can actually consider $\sum_{n=0}^{999} \cos^2(n\pmod{\pi})$ . However, when $N$ is big, the numbers $n\pmod{\pi}$ for $n=0, 1, \ldots, N-1$ "look like" $N$ random variables which are uniform $[0,\pi)$ . However, then, if we write $X \sim U[0,\pi)$ , $\cos^2\frac{\pi}{\alpha} = \frac{1}{1000}\sum_{n=0}^{999} \cos^2n \approx \mathbb{E}[\cos^2X] = \mathbb{E}\left[\frac{1}{2}+\frac{1}{2}\cos(2X)\right] = \frac{1}{2}.$

Then $\cos\frac{\pi}{\alpha} \approx \pm\frac{\sqrt{2}}{2}$ $\frac{\pi}{\alpha} \approx \frac{\pi}{4} + \frac{k\pi}{2}$ $\alpha \approx \frac{4}{1+2k}$

Setting $k=0$ yields $\alpha \approx \boxed{4}$ .

Note that the given sum equals $(1/2)(1000+1/2+1/2(sin1999/sin1))$ .This follows from the summation $cosx+cos(x+y)+cos(x+2y)+...........cos(x+(n-1)y)=cos(x+(n-1)y/2)Sin(ny/2)/Sin(y/2)$ .Now The given sum is approximately equal to $500+b$ where $b<1$ .Hence $cos(π/\alpha$ )) is almost $1/√$ 2....So the required value is $\alpha$ = $π/4$