Let $n\geq2$ be a positive integer. Determine all possible values of the sum
\[ S=\left\lfloor x_{2}-x_{1}\right\rfloor +\left\lfloor x_{3}-x_{2}\right\rfloor+...+\left\lfloor x_{n}-x_{n-1}\right\rfloor \]
 where $x_i\in \mathbb{R}$ satisfying $\lfloor{x_i}\rfloor=i$ for $i=1,2,\ldots n$.
1. **Understanding the Problem:**
   We need to determine all possible values of the sum 
   \[
   S = \left\lfloor x_{2} - x_{1} \right\rfloor + \left\lfloor x_{3} - x_{2} \right\rfloor + \cdots + \left\lfloor x_{n} - x_{n-1} \right\rfloor
   \]
   where \( x_i \in \mathbb{R} \) and \( \lfloor x_i \rfloor = i \) for \( i = 1, 2, \ldots, n \).

2. **Analyzing the Constraints:**
   Since \( \lfloor x_i \rfloor = i \), we have \( i \leq x_i < i+1 \) for each \( i \). This implies:
   \[
   x_1 \in [1, 2), \quad x_2 \in [2, 3), \quad \ldots, \quad x_n \in [n, n+1)
   \]

3. **Non-negativity of Summands:**
   For each \( i \), \( x_{i+1} \geq i+1 \) and \( x_i < i+1 \), so \( x_{i+1} > x_i \). Therefore, \( x_{i+1} - x_i \geq 1 \), and thus:
   \[
   \left\lfloor x_{i+1} - x_i \right\rfloor \geq 0
   \]
   This shows that all summands are non-negative.

4. **Upper Bound of the Sum:**
   Consider the sum of the differences:
   \[
   (x_2 - x_1) + (x_3 - x_2) + \cdots + (x_n - x_{n-1}) = x_n - x_1
   \]
   Since \( x_1 \in [1, 2) \) and \( x_n \in [n, n+1) \), we have:
   \[
   x_n - x_1 < (n+1) - 1 = n
   \]
   Therefore, the sum \( S \) does not exceed \( n-1 \).

5. **Constructing Examples:**
   To show that all values from \( 0 \) to \( n-1 \) are possible, consider the following construction:
   - Let \( k \) be an integer such that \( 0 \leq k \leq n-1 \).
   - Define \( x_i \) as follows:
     \[
     x_i = i + a_i \quad \text{where} \quad a_i \in [0, 1)
     \]
     Specifically, choose \( a_i \) such that:
     \[
     a_1 = 0, \quad a_2 = 0, \quad \ldots, \quad a_{n-k} = 0, \quad a_{n-k+1} = 1 - \epsilon, \quad a_{n-k+2} = 1 - \epsilon, \quad \ldots, \quad a_n = 1 - \epsilon
     \]
     where \( \epsilon \) is a small positive number.

6. **Calculating the Sum:**
   For this construction:
   \[
   \left\lfloor x_{i+1} - x_i \right\rfloor = 
   \begin{cases} 
   0 & \text{for } i = 1, 2, \ldots, n-k-1 \\
   1 & \text{for } i = n-k, n-k+1, \ldots, n-1 
   \end{cases}
   \]
   Therefore, the sum \( S \) is:
   \[
   S = 0 + 0 + \cdots + 0 + 1 + 1 + \cdots + 1 = k
   \]
   This shows that \( S \) can take any value from \( 0 \) to \( n-1 \).

Conclusion:
The possible values of \( S \) are \( 0, 1, 2, \ldots, n-1 \).

The final answer is \(\boxed{0, 1, 2, \ldots, n-1}\).