Let $x_{2i} = 0, x_{2i-1} = \frac{1}{2}$ for all $i = 1, \ldots, 50$. Then we have $S = 50 \cdot \left( \frac{1}{2} \right)^2 = \frac{25}{2}$. So, we are left to show that $S \leq \frac{25}{2}$ for all values of $x_i$'s satisfying the problem conditions. Consider any $1 \leq i \leq 50$. By the problem condition, we get $x_{2i-1} \leq 1 - x_{2i} - x_{2i+1}$ and $x_{2i+2} \leq 1 - x_{2i} - x_{2i+1}$. Hence by the AM-GM inequality we get
$$\begin{aligned} x_{2i-1} x_{2i+1} + x_{2i} x_{2i+2} & \leq \left( 1 - x_{2i} - x_{2i+1} \right) x_{2i+1} + x_{2i} \left( 1 - x_{2i} - x_{2i+1} \right) \\ & = \left( x_{2i} + x_{2i+1} \right) \left( 1 - x_{2i} - x_{2i+1} \right) \leq \left( \frac{ \left( x_{2i} + x_{2i+1} \right) + \left( 1 - x_{2i} - x_{2i+1} \right) }{2} \right)^2 = \frac{1}{4} \end{aligned}$$
Summing up these inequalities for $i = 1, 2, \ldots, 50$, we get the desired inequality
$$\sum_{i=1}^{50} \left( x_{2i-1} x_{2i+1} + x_{2i} x_{2i+2} \right) \leq 50 \cdot \frac{1}{4} = \frac{25}{2}$$
Comment. This solution shows that a bit more general fact holds. Namely, consider $2n$ nonnegative numbers $x_1, \ldots, x_{2n}$ in a row (with no cyclic notation) and suppose that $x_i + x_{i+1} + x_{i+2} \leq 1$ for all $i = 1, 2, \ldots, 2n-2$. Then $\sum_{i=1}^{2n-2} x_i x_{i+2} \leq \frac{n-1}{4}$. The proof is the same as above, though if might be easier to find it (for instance, applying induction). The original estimate can be obtained from this version by considering the sequence $x_1, x_2, \ldots, x_{100}, x_1, x_2$.