Let $ y_i = \frac {x_i}{2}$(if $ i$=odd)and $ y_i = 2x_i$(if $ i$ is even),the system is equivalent to $ y_1 + \frac {1}{y_2} = 2, y_2 + \frac {1}{y_3} = 2, y_3 + \frac {1}{y_4} = 2, ..., y_{99} + \frac {1}{y_{100}} = 2, y_{100} + \frac {1}{y_1} = 2$. Sum all each hands,we have $ (y_1 + \frac {1}{y_1}) + (y_2 + \frac {1}{y_2}) + \cdots + (y_{99} + \frac {1}{y_{99}}) + (y_{100} + \frac {1}{y_{100}}) = 200$. By AM-GM inequality, $ y_i + \frac {1}{y_i}\ge 2$ holds for all $ i$,thus we have $ (y_1 + \frac {1}{y_1}) + (y_2 + \frac {1}{y_2}) + \cdots + (y_{99} + \frac {1}{y_{99}}) + (y_{100} + \frac {1}{y_{100}})\ge 200$ therefore,the equality of the AM-GM must holds for all $ i$,so we have $ y_1 = y_2 = \cdots = y_{100} = 1$ and we see the solution $ \boxed{x_1 = x_3 = \cdots = x_{99} = 2, x_2 = x_4 = \cdots = x_{100} = \frac {1}{2}}$.