Given the constraints and objective of the problem, we aim to find the maximal possible value of the sum $S = \sum_{i=1}^{100} x_i x_{i+2}$ where the sequence $x_1, \ldots, x_{100}$ consists of nonnegative real numbers satisfying the condition:

$$x_i + x_{i+1} + x_{i+2} \leq 1 \quad \text{for all } i = 1, \ldots, 100.$$

Here, indices are cyclic, so $x_{101} = x_1$ and $x_{102} = x_2$.

### Step-by-step Solution:

1. Understanding the Constraint:

The key constraint is:

$$x_i + x_{i+1} + x_{i+2} \leq 1.$$

This condition must hold for each subsequent triplet in the sequence, creating a cyclic condition for 100 terms.

2. Approach to Solve:

We adopt a strategy using periodic patterns due to symmetry and cycle:

- For simplicity, assume a repeating pattern cycle of three consecutive numbers: $x_i, x_{i+1}, x_{i+2} = a, b, c$.

With the given constraint:
$$a + b + c \leq 1.$$

- Using symmetry, set $x_i = x_{i+3} = x_{i+6} = \ldots$ repeating sequences of the form $[ a, 0, b, 0, c, 0 ]$.

Each computation of $x_i \cdot x_{i+2}$ simplifies due to the zero elements in the repeated sequence yielding:
$$S = \sum_{j=0}^{33} (x_{3j+1} \cdot x_{3j+3} + x_{3j+2} \cdot x_{3j+4} + x_{3j+3} \cdot x_{3j+5}).$$

**3. Maximizing the Sum $S$:**

- For simplicity, assume $x_{3j+3} = a, x_{3j+5} = b, \text{and} \ x_{3j+1} = x_{3j+4} = c$.

Then, you can express it as:
- Each pair $(x_i, x_{i+2})$ meets once:
$$x_i x_{i+2} = a \cdot b + b \cdot c + c \cdot a$$

The goal is to maximize the total over these combinations utilizing $a + b + c \leq 1$. The largest achievable for each cycle:

$$a = b = c = \frac{1}{2},$$

resulting in:

$$x_i \cdot x_{i+2} = \frac{1}{4}.$$

Each cycle is repeated oscillating over 100 indices, yielding the maximal sum:

$$S \rightarrow \frac{100}{4} = \frac{25}{2}.$$

The answer confirms the maximum possible sum of product pairs is then:

$$\boxed{\frac{25}{2}}.$$

Thus, the maximal possible value of $S = \sum_{i=1}^{100} x_i x_{i+2}$ is $\boxed{\frac{25}{2}}$.