Find all positive integers $n$ such that for all real numbers $x_1,x_2,\ldots ,x_n,y_1,y_2,\ldots ,y_n$ the following inequality holds:
\[ x_1x_2\ldots x_n+y_1y_2\ldots y_n\le\sqrt{x_1^2+y_1^2}\cdot\sqrt{x_2^2+y_2^2}\cdot \cdots \sqrt{x_n^2+y_n^2}\cdot  \]
To find all positive integers \( n \) such that for all real numbers \( x_1, x_2, \ldots, x_n, y_1, y_2, \ldots, y_n \), the following inequality holds:
\[ x_1 x_2 \ldots x_n + y_1 y_2 \ldots y_n \le \sqrt{x_1^2 + y_1^2} \cdot \sqrt{x_2^2 + y_2^2} \cdot \cdots \cdot \sqrt{x_n^2 + y_n^2}, \]
we will use the Cauchy-Schwarz inequality and Holder's inequality.

1. **Base Case \( n = 2 \):**
   For \( n = 2 \), we can use the Cauchy-Schwarz inequality in \(\mathbb{R}^2\):
   \[
   (x_1 y_2 - x_2 y_1)^2 \ge 0 \implies x_1^2 y_2^2 + x_2^2 y_1^2 \ge 2 x_1 x_2 y_1 y_2.
   \]
   Adding \( x_1^2 y_1^2 + x_2^2 y_2^2 \) to both sides, we get:
   \[
   x_1^2 y_1^2 + x_2^2 y_2^2 + x_1^2 y_2^2 + x_2^2 y_1^2 \ge x_1^2 y_1^2 + x_2^2 y_2^2 + 2 x_1 x_2 y_1 y_2.
   \]
   This simplifies to:
   \[
   (x_1^2 + y_1^2)(x_2^2 + y_2^2) \ge (x_1 x_2 + y_1 y_2)^2.
   \]
   Taking the square root of both sides, we obtain:
   \[
   \sqrt{x_1^2 + y_1^2} \cdot \sqrt{x_2^2 + y_2^2} \ge x_1 x_2 + y_1 y_2.
   \]
   Thus, the inequality holds for \( n = 2 \).

2. **General Case \( n \ge 3 \):**
   We need to prove:
   \[
   x_1 x_2 \ldots x_n + y_1 y_2 \ldots y_n \le \sqrt{x_1^2 + y_1^2} \cdot \sqrt{x_2^2 + y_2^2} \cdot \cdots \cdot \sqrt{x_n^2 + y_n^2}.
   \]
   Using Holder's inequality, we have:
   \[
   \left( \sum_{i=1}^n a_i b_i \right)^n \le \left( \sum_{i=1}^n a_i^n \right) \left( \sum_{i=1}^n b_i^n \right),
   \]
   where \( a_i = \sqrt[n]{x_i^2 + y_i^2} \) and \( b_i = 1 \). Applying this to our problem, we get:
   \[
   (x_1 x_2 \ldots x_n + y_1 y_2 \ldots y_n)^n \le \left( \sum_{i=1}^n \sqrt[n]{x_i^2 + y_i^2} \right)^n.
   \]
   Simplifying, we have:
   \[
   x_1 x_2 \ldots x_n + y_1 y_2 \ldots y_n \le \sqrt[n]{(x_1^2 + y_1^2)(x_2^2 + y_2^2) \ldots (x_n^2 + y_n^2)}.
   \]
   Since \( \sqrt[n]{(x_1^2 + y_1^2)(x_2^2 + y_2^2) \ldots (x_n^2 + y_n^2)} = \sqrt{x_1^2 + y_1^2} \cdot \sqrt{x_2^2 + y_2^2} \cdot \cdots \cdot \sqrt{x_n^2 + y_n^2} \), the inequality holds for all \( n \ge 2 \).

3. **Conclusion:**
   The inequality holds for all \( n \ge 2 \). For \( n = 1 \), the inequality does not hold as it reduces to \( x_1 + y_1 \le \sqrt{x_1^2 + y_1^2} \), which is not true for all real numbers \( x_1 \) and \( y_1 \).

The final answer is \( \boxed{ n \ge 2 } \).