By inspection, $ (\sqrt{17},\sqrt{17},\sqrt{17},\sqrt{17}) $ and $ (-\sqrt{17},-\sqrt{17},-\sqrt{17},-\sqrt{17}) $ are solutions. We claim that these are the only solutions. Let \[f(x) = \frac{1}{2} \left( x + \frac{17}{x} \right) = \frac{x^2 + 17}{2x}.\]Then the given equations become $ f(x) = y, $ $ f(y) = z, $ $ f(z) = w, $ and $ f(w) = x $. Note that none of these variables can be 0. Suppose $ t > 0 $. Then \[f(t) - \sqrt{17} = \frac{t^2 + 17}{2t} - \sqrt{17} = \frac{t^2 - 2t \sqrt{17} + 17}{2t} = \frac{(t - \sqrt{17})^2}{2t} \ge 0,\]so $ f(t) \ge \sqrt{17} $. Hence, if any of $ x, $ $ y, $ $ z, $ $ w $ are positive, then they are all positive, and greater than or equal to $ \sqrt{17} $. Furthermore, if $ t > \sqrt{17}, $ then \[f(t) - \sqrt{17} = \frac{(t - \sqrt{17})^2}{2t} = \frac{1}{2} \cdot \frac{t - \sqrt{17}}{t} (t - \sqrt{17}) < \frac{1}{2} (t - \sqrt{17}).\]Hence, if $ x > \sqrt{17}, $ then \begin{align*} y - \sqrt{17} &< \frac{1}{2} (x - \sqrt{17}), \\ z - \sqrt{17} &< \frac{1}{2} (y - \sqrt{17}), \\ w - \sqrt{17} &< \frac{1}{2} (z - \sqrt{17}), \\ x - \sqrt{17} &< \frac{1}{2} (w - \sqrt{17}).\end{align*}This means \[x - \sqrt{17} < \frac{1}{2} (w - \sqrt{17}) < \frac{1}{4} (z - \sqrt{17}) < \frac{1}{8} (y - \sqrt{17}) < \frac{1}{16} (x - \sqrt{17}),\]contradiction. Therefore, $ (\sqrt{17},\sqrt{17},\sqrt{17},\sqrt{17}) $ is the only solution where any of the variables are positive. If any of the variables are negative, then they are all negative. Let $ x' = -x, $ $ y' = -y, $ $ z' = -z, $ and $ w' = -w $. Then \begin{align*} 2y' &= x' + \frac{17}{x'}, \\ 2z' &= y' + \frac{17}{y'}, \\ 2w' &= z' + \frac{17}{z'}, \\ 2x' &= w' + \frac{17}{w'}, \end{align*}and $ x', $ $ y', $ $ z', $ $ w' $ are all positive, which means $ (x',y',z',w') = (\sqrt{17},\sqrt{17},\sqrt{17},\sqrt{17}), $ so $ (x,y,z,w) = (-\sqrt{17},-\sqrt{17},-\sqrt{17},-\sqrt{17}) $. Thus, there are $ \boxed{2} $ solutions.