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Problem Statement:

We are given a sequence of numbers \( x_1, x_2, \ldots, x_n \) such that the sum of their squares is 1, i.e.,

\[
x_1^2 + x_2^2 + \cdots + x_n^2 = 1.
\]

We need to find the maximum value of the expression:

\[
S = \sum_{i=1}^{n-1} x_i x_{i+1}.
\]

Solution Approach:

#### Step 1: Understanding the Expression
The expression \( S = \sum_{i=1}^{n-1} x_i x_{i+1} \) represents the sum of the products of consecutive terms in the sequence. Our goal is to maximize this sum under the constraint \( x_1^2 + x_2^2 + \cdots + x_n^2 = 1 \).

#### Step 2: Using the Cauchy-Schwarz Inequality
To find an upper bound for \( S \), we can use the Cauchy-Schwarz inequality. Consider the vectors \( \mathbf{a} = (x_1, x_2, \ldots, x_{n-1}) \) and \( \mathbf{b} = (x_2, x_3, \ldots, x_n) \). The Cauchy-Schwarz inequality states:

\[
\left( \sum_{i=1}^{n-1} x_i x_{i+1} \right)^2 \leq \left( \sum_{i=1}^{n-1} x_i^2 \right) \left( \sum_{i=2}^{n} x_i^2 \right).
\]

Let us denote:

\[
A = \sum_{i=1}^{n-1} x_i^2 \quad \text{and} \quad B = \sum_{i=2}^{n} x_i^2.
\]

Then, the inequality becomes:

\[
S^2 \leq A \cdot B.
\]

#### Step 3: Relating \( A \) and \( B \)
From the given constraint \( x_1^2 + x_2^2 + \cdots + x_n^2 = 1 \), we have:

\[
A + x_n^2 = 1 \quad \text{and} \quad B + x_1^2 = 1.
\]

Thus,

\[
A = 1 - x_n^2 \quad \text{and} \quad B = 1 - x_1^2.
\]

Substituting these into the inequality, we get:

\[
S^2 \leq (1 - x_n^2)(1 - x_1^2).
\]

Expanding the right-hand side:

\[
(1 - x_n^2)(1 - x_1^2) = 1 - x_1^2 - x_n^2 + x_1^2 x_n^2.
\]

#### Step 4: Maximizing \( S \)
To maximize \( S \), we need to consider the equality case of the Cauchy-Schwarz inequality, which occurs when the vectors \( \mathbf{a} \) and \( \mathbf{b} \) are proportional. This implies:

\[
x_1 = x_2 = \cdots = x_n.
\]

However, since \( x_1^2 + x_2^2 + \cdots + x_n^2 = 1 \), if all \( x_i \) are equal, then:

\[
x_1 = x_2 = \cdots = x_n = \frac{1}{\sqrt{n}}.
\]

Substituting \( x_i = \frac{1}{\sqrt{n}} \) into the expression for \( S \):

\[
S = \sum_{i=1}^{n-1} x_i x_{i+1} = \sum_{i=1}^{n-1} \left( \frac{1}{\sqrt{n}} \right) \left( \frac{1}{\sqrt{n}} \right) = \sum_{i=1}^{n-1} \frac{1}{n} = \frac{n-1}{n}.
\]

#### Step 5: Verifying the Maximum Value
The value \( S = \frac{n-1}{n} \) is achievable when \( x_1 = x_2 = \cdots = x_n = \frac{1}{\sqrt{n}} \). Therefore, the maximum value of \( S \) is:

\[
\boxed{\frac{n-1}{2}}.
\]

This result is consistent with the problem's constraints and the application of the Cauchy-Schwarz inequality. The final answer is:

\[
\boxed{\frac{n-1}{2}}.
\]