Let $n$ be a positive integer. Find the maximal real number $k$, such that the following holds:

For any $n$ real numbers $x_1,x_2,...,x_n$, we have $\sqrt{x_1^2+x_2^2+\dots+x_n^2}\geq k\cdot\min(|x_1-x_2|,|x_2-x_3|,...,|x_{n-1}-x_n|,|x_n-x_1|)$
1. **Consider the problem for \( n > 1 \)**:
   We need to find the maximal real number \( k \) such that for any \( n \) real numbers \( x_1, x_2, \ldots, x_n \), the inequality 
   \[
   \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2} \geq k \cdot \min(|x_1 - x_2|, |x_2 - x_3|, \ldots, |x_{n-1} - x_n|, |x_n - x_1|)
   \]
   holds.

2. **Claim**:
   We claim that the answer is \( \frac{\sqrt{n}}{2} \) if \( n \) is even and \( \frac{\sqrt{n+5}}{2} \) if \( n \) is odd.

3. **Verification for even \( n \)**:
   - Consider the sequence \( (x_1, x_2, \ldots, x_n) = (1, -1, 1, -1, \ldots, 1, -1) \).
   - For this sequence, we have:
     \[
     x_1^2 + x_2^2 + \cdots + x_n^2 = n
     \]
     and
     \[
     \min(|x_1 - x_2|, |x_2 - x_3|, \ldots, |x_{n-1} - x_n|, |x_n - x_1|) = 2
     \]
   - Therefore, the inequality becomes:
     \[
     \sqrt{n} \geq k \cdot 2 \implies k \leq \frac{\sqrt{n}}{2}
     \]
   - To show that \( k = \frac{\sqrt{n}}{2} \) is achievable, we use the inequality:
     \[
     x_i^2 + x_{i+1}^2 \geq \frac{1}{2} (x_i - x_{i+1})^2
     \]
     Summing this over \( i = 1, 3, 5, \ldots, n-1 \) yields:
     \[
     x_1^2 + x_2^2 + \cdots + x_n^2 \geq \frac{1}{2} \sum_{i=1,3,5,\ldots,n-1} (x_i - x_{i+1})^2
     \]
     Since there are \( \frac{n}{2} \) terms, we get:
     \[
     x_1^2 + x_2^2 + \cdots + x_n^2 \geq \frac{n}{2} \cdot 4 = 2n
     \]
     This confirms that \( k = \frac{\sqrt{n}}{2} \) is achievable.

4. **Verification for odd \( n \)**:
   - Consider the sequence \( (x_1, x_2, \ldots, x_n) = (2, 0, -2, 1, -1, 1, -1, \ldots, 1, -1) \).
   - For this sequence, we have:
     \[
     x_1^2 + x_2^2 + \cdots + x_n^2 = n + 5
     \]
     and
     \[
     \min(|x_1 - x_2|, |x_2 - x_3|, \ldots, |x_{n-1} - x_n|, |x_n - x_1|) = 2
     \]
   - Therefore, the inequality becomes:
     \[
     \sqrt{n+5} \geq k \cdot 2 \implies k \leq \frac{\sqrt{n+5}}{2}
     \]
   - To show that \( k = \frac{\sqrt{n+5}}{2} \) is achievable, we use the inequality:
     \[
     x_i^2 + x_{i+1}^2 + x_{i+2}^2 \geq x_i^2 + x_{i+2}^2 \geq \frac{1}{2} (x_i - x_{i+2})^2 \geq 2 (x_{i+1} - x_i)(x_{i+2} - x_{i+1})
     \]
     Summing with \( \frac{n-3}{2} \) other inequalities of the form \( x_j^2 + x_{j+1}^2 \geq \frac{1}{2} (x_j - x_{j+1})^2 \) yields the result.

The final answer is \( \boxed{\frac{\sqrt{n}}{2} \text{ if } n \text{ is even and } \frac{\sqrt{n+5}}{2} \text{ if } n \text{ is odd}} \).