[THOUGHT] To solve the problem, we need to maximize the sum $S = \sum_{i=1}^{n} x_i x_{i+2}$ under the constraint that $x_i + x_{i+1} + x_{i+2} \leq 1$ for all $i$, with indices taken modulo $n$. For even $n$, a two-periodic pattern where variables alternate between $a$ and $b$ is optimal. This pattern satisfies the constraints $2a + b \leq 1$ and $a + 2b \leq 1$. The sum $S$ becomes $\frac{n}{2}(a^2 + b^2)$. The maximum value of $a^2 + b^2$ under these constraints is $\frac{1}{4}$, achieved when $a = \frac{1}{2}$ and $b = 0$ (or vice versa). Thus, $S = \frac{n}{8}$.

For $n = 2026$, the maximum value $M_{2026} = \frac{2026}{8} = \frac{1013}{4}$. Since 1013 is a prime number, the fraction is in lowest terms. Therefore, $p = 1013$ and $q = 4$, giving $p + q = 1017$. [/THOUGHT]

[PYTHON]

Verify the calculation for n=2026

n = 2026 p = n // 2 # Since 2026 / 8 = 1013 / 4, p=1013, q=4 q = 4

Check if 1013 and 4 are coprime

from math import gcd assert gcd(p, q) == 1, "p and q are not coprime" print(p + q) [/PYTHON]

[THOUGHT] The Python code confirms that $p = 1013$ and $q = 4$ are coprime, leading to $p + q = 1017$. This matches the derived result for $M_{2026} = \frac{1013}{4}$. [/THOUGHT] $\boxed{1017}$