Let $n\geq 3$ be an integer. Find the largest real number $M$ such that for any positive real numbers $x_1,x_2,\cdots,x_n$, there exists an arrangement $y_1,y_2,\cdots,y_n$ of real numbers satisfying
\[\sum_{i=1}^n \frac{y_i^2}{y_{i+1}^2-y_{i+1}y_{i+2}+y_{i+2}^2}\geq M,\]
where $y_{n+1}=y_1,y_{n+2}=y_2$.
1. **Initial Setup and Assumptions:**
   Let \( n \geq 3 \) be an integer. We need to find the largest real number \( M \) such that for any positive real numbers \( x_1, x_2, \ldots, x_n \), there exists an arrangement \( y_1, y_2, \ldots, y_n \) of these numbers satisfying:
   \[
   \sum_{i=1}^n \frac{y_i^2}{y_{i+1}^2 - y_{i+1} y_{i+2} + y_{i+2}^2} \geq M,
   \]
   where \( y_{n+1} = y_1 \) and \( y_{n+2} = y_2 \).

2. **Choosing an Arrangement:**
   Assume \( y_1 \geq y_2 \geq \cdots \geq y_n \). This arrangement helps in simplifying the inequality.

3. **Simplifying the Sum:**
   Consider the sum:
   \[
   \sum_{i=1}^{n} \frac{y_i^2}{y_{i+1}^2 - y_{i+1} y_{i+2} + y_{i+2}^2}.
   \]
   We can rewrite the sum by considering the cyclic nature of the indices:
   \[
   \frac{y_{n-1}^2}{y_n^2 - y_n y_1 + y_1^2} + \sum_{i=1}^{n-2} \frac{y_i^2}{y_{i+1}^2 + y_{i+2}(y_{i+2} - y_{i+1})}.
   \]

4. **Bounding the Terms:**
   Notice that:
   \[
   \frac{y_i^2}{y_{i+1}^2 - y_{i+1} y_{i+2} + y_{i+2}^2} > \frac{y_i^2}{y_{i+1}^2}.
   \]
   Therefore, we have:
   \[
   \sum_{i=1}^{n} \frac{y_i^2}{y_{i+1}^2 - y_{i+1} y_{i+2} + y_{i+2}^2} > \sum_{i=1}^{n} \frac{y_i^2}{y_{i+1}^2}.
   \]

5. **Using the Rearrangement Inequality:**
   By the rearrangement inequality, we know that:
   \[
   \sum_{i=1}^{n} \frac{y_i^2}{y_{i+1}^2} \geq n-1.
   \]
   This is because the sum of the ratios of the squares of the terms in a decreasing sequence is minimized when the terms are arranged in the same order.

6. **Conclusion:**
   Therefore, we have:
   \[
   \sum_{i=1}^{n} \frac{y_i^2}{y_{i+1}^2 - y_{i+1} y_{i+2} + y_{i+2}^2} \geq n-1.
   \]
   Furthermore, when \( y_1 = y_2 = \cdots = y_{n-1} = 1 \) and \( y_n \) approaches 0, the left-hand side approaches \( n-1 \) as well. Hence, the largest \( M \) is \( n-1 \).

The final answer is \( \boxed{ n-1 } \).