Answer: 582

Solution:

By rotational symmetry we may fix $v$ and let $w = v e^{2\pi i k/1997}$ with $k$ uniform in $\{1, 2, \ldots, 1996\}.$ Then $$\begin{aligned} |v + w| &= \left|1 + e^{2\pi i k/1997}\right| \\ &= 2\left|\cos\frac{\pi k}{1997}\right|. \end{aligned}$$ Also $\left(2\cos 15^\circ\right)^2 = 2 + 2\cos 30^\circ$ $= 2 + \sqrt{3},$ so the threshold is $\sqrt{2 + \sqrt{3}} = 2\cos\frac{\pi}{12}.$

The condition $\left|\cos\frac{\pi k}{1997}\right| \ge \cos\frac{\pi}{12}$ holds exactly when $\frac{\pi k}{1997}$ is within $\frac{\pi}{12}$ of $0$ or of $\pi,$ i.e. $k \le \frac{1997}{12} = 166.41\ldots$ or $k \ge \frac{11 \cdot 1997}{12} = 1830.58\ldots.$ That gives $166 + 166 = 332$ favorable values of $k.$

The probability is $\frac{332}{1996} = \frac{83}{499},$ and $499$ is prime, so $m + n = 83 + 499 = 582.$