Let $ n \geq 3$ be an odd integer. Determine the maximum value of
\[ \sqrt{|x_{1}\minus{}x_{2}|}\plus{}\sqrt{|x_{2}\minus{}x_{3}|}\plus{}\ldots\plus{}\sqrt{|x_{n\minus{}1}\minus{}x_{n}|}\plus{}\sqrt{|x_{n}\minus{}x_{1}|},\]
where $ x_{i}$ are positive real numbers from the interval $ [0,1]$.
To determine the maximum value of the given expression, we need to analyze the function:

\[
S = \sqrt{|x_1 - x_2|} + \sqrt{|x_2 - x_3|} + \ldots + \sqrt{|x_{n-1} - x_n|} + \sqrt{|x_n - x_1|}
\]

where \( x_i \) are positive real numbers from the interval \([0,1]\) and \( n \geq 3 \) is an odd integer.

1. **Existence of Optimal Point**:
   Since the function is continuous on the compact set \([0,1]^n\), there exists an optimal point \((x_1, x_2, \ldots, x_n)\) that maximizes \(S\).

2. **Characterization of Optimal Points**:
   We analyze the conditions under which the function \(S\) can be maximized:
   - If \(x_i \leq x_{i-1}\) and \(x_i \leq x_{i+1}\), then \(x_i = 0\).
   - If \(x_i \geq x_{i-1}\) and \(x_i \geq x_{i+1}\), then \(x_i = 1\).
   - If \(x_{i+1} \leq x_i \leq x_{i-1}\) or \(x_{i-1} \leq x_i \leq x_{i+1}\), then \(x_i = \frac{x_{i-1} + x_{i+1}}{2}\).

3. **Structure of Optimal Points**:
   Based on the above conditions, the sequence \((x_1, x_2, \ldots, x_n)\) can be structured as follows:
   \[
   (0, \frac{1}{k_1}, \frac{2}{k_1}, \ldots, 1, \frac{k_2 - 1}{k_2}, \ldots, \frac{2}{k_2}, \frac{1}{k_2}, 0, \frac{1}{k_3}, \ldots, \frac{1}{k_l})
   \]
   where \(k_1, k_2, \ldots, k_l\) are natural numbers such that \(k_1 + k_2 + \ldots + k_l = n\) and \(l\) is even.

4. **Expression for \(S\)**:
   The function \(S\) at this point equals:
   \[
   S = \sqrt{k_1} + \sqrt{k_2} + \ldots + \sqrt{k_l}
   \]

5. **Maximizing \(S\)**:
   To maximize \(S\), we use the fact that \(\sqrt{k} < \sqrt{k-1} + 1\). Given that \(l\) is even, the maximum possible value of \(S\) is achieved when \(l = n - 1\), \(k_1 = k_2 = \ldots = k_{l-1} = 1\), and \(k_l = 2\). Thus, we have:
   \[
   S = (n - 2) \cdot \sqrt{1} + \sqrt{2} = n - 2 + \sqrt{2}
   \]

The final answer is \(\boxed{n - 2 + \sqrt{2}}\).