Answer: $\frac{25}{2}$.
Let $x_{2i} = 0, x_{2i-1} = \frac{1}{2}$ for all $i = 1, \cdots, 50$. Then we have
$$S = 50\left(\frac{1}{2}\right)^2 = \frac{25}{2}.$$
Thus it suffices to show $S \le \frac{25}{2}$ for all $x_i$ satisfying the given conditions.
Consider $1 \le i \le 50$. From the given conditions we obtain
$$x_{2i-1} \le 1 - x_{2i} - x_{2i+1}, \quad x_{2i+2} \le 1 - x_{2i} - x_{2i+1}.$$
Then by the AM-GM inequality we get
$$\begin{aligned} x_{2i-1}x_{2i+1} + x_2x_{2i+2} &\le (1-x_{2i}-x_{2i+1})x_{2i+1} + x_2(1-x_{2i}-x_{2i+1}) \\ &= (x_{2i}+x_{2i+1})(1-x_{2i}-x_{2i+1}) \\ &\le \left(\frac{x_{2i}+x_{2i+1}}{2} + (1-x_{2i}-x_{2i+1})\right)^2 = \frac{1}{4}. \end{aligned}$$
Summing the above inequality for $i = 1, \cdots, 50$, we obtain
$$\sum_{i=1}^{50} (x_{2i-1}x_{2i+1} + x_2x_{2i+2}) \le 50 \cdot \frac{1}{4} = \frac{25}{2}.$$