For a given integer $n\geq 3$, determine the range of values for the expression
\[ E_n(x_1,x_2,\ldots,x_n) := \dfrac {x_1} {x_2} + \dfrac {x_2} {x_3} + \cdots + \dfrac {x_{n-1}} {x_n} + \dfrac {x_n} {x_1}\]
over real numbers $x_1,x_2,\ldots,x_n \geq 1$ satisfying $|x_k - x_{k+1}| \leq 1$ for all $1\leq k \leq n-1$.
Do also determine when the extremal values are achieved.

(Dan Schwarz)
1. **Identify the minimum value of \( E_n \):**
   \[
   E_n(x_1, x_2, \ldots, x_n) = \frac{x_1}{x_2} + \frac{x_2}{x_3} + \cdots + \frac{x_{n-1}}{x_n} + \frac{x_n}{x_1}
   \]
   When \( x_1 = x_2 = \cdots = x_n \), each term in the sum is equal to 1. Therefore,
   \[
   E_n = n
   \]
   This is the minimum value of \( E_n \).

2. **Prove the upper bound using induction:**
   We aim to show that:
   \[
   E_n \leq \frac{1}{2} + \frac{2}{3} + \cdots + \frac{n-1}{n} + n
   \]
   for any \( n \geq 2 \).

3. **Base case \( n = 2 \):**
   \[
   E_2 = \frac{x_1}{x_2} + \frac{x_2}{x_1}
   \]
   We need to show:
   \[
   \frac{x_1}{x_2} + \frac{x_2}{x_1} \leq \frac{5}{2}
   \]
   This can be rewritten as:
   \[
   \frac{(2x_1 - x_2)(2x_2 - x_1)}{x_1 x_2} \geq 0
   \]
   Since \( x_1, x_2 \geq 1 \) and \( |x_1 - x_2| \leq 1 \), we have:
   \[
   2x_1 - x_2 \geq 1 + x_1 - x_2 \geq 0 \quad \text{and} \quad 2x_2 - x_1 \geq 1 + x_2 - x_1 \geq 0
   \]
   Thus, the base case holds.

4. **Inductive step:**
   Assume the inequality holds for \( n-1 \):
   \[
   E_{n-1} \leq \frac{1}{2} + \frac{2}{3} + \cdots + \frac{n-2}{n-1} + (n-1)
   \]
   We need to show:
   \[
   E_n \leq \frac{1}{2} + \frac{2}{3} + \cdots + \frac{n-1}{n} + n
   \]
   Using the induction hypothesis and the inequality:
   \[
   \frac{x_{n-1}}{x_n} + \frac{x_n}{x_1} \leq \frac{x_{n-1}}{x_1} + \frac{2n-1}{n}
   \]
   We have:
   \[
   E_n = E_{n-1} + \frac{x_{n-1}}{x_n} + \frac{x_n}{x_1} - \frac{x_{n-1}}{x_1}
   \]
   Substituting the induction hypothesis and the inequality:
   \[
   E_n \leq \left( \frac{1}{2} + \frac{2}{3} + \cdots + \frac{n-2}{n-1} + (n-1) \right) + \frac{2n-1}{n}
   \]
   Simplifying, we get:
   \[
   E_n \leq \frac{1}{2} + \frac{2}{3} + \cdots + \frac{n-1}{n} + n
   \]
   Thus, the inductive step holds.

5. **Conclusion:**
   We have shown that:
   \[
   n \leq E_n \leq \frac{1}{2} + \frac{2}{3} + \cdots + \frac{n-1}{n} + n
   \]
   The minimum value \( E_n = n \) is achieved when \( x_1 = x_2 = \cdots = x_n \).

The final answer is \( \boxed{ n \leq E_n \leq \frac{1}{2} + \frac{2}{3} + \cdots + \frac{n-1}{n} + n } \)