Let $n\geq 2$ an integer. Find the least value of $\gamma$ such that for any positive real numbers $x_1,x_2,...,x_n$ with $x_1+x_2+...+x_n=1$ and any real $y_1+y_2+...+y_n=1$ and $0\leq y_1,y_2,...,y_n\leq \frac{1}{2}$ the following inequality holds:
$$x_1x_2...x_n\leq \gamma \left(x_1y_1+x_2y_2+...+x_ny_n\right)$$
To find the least value of $\gamma$ such that for any positive real numbers $x_1, x_2, \ldots, x_n$ with $x_1 + x_2 + \cdots + x_n = 1$ and any real numbers $y_1, y_2, \ldots, y_n$ with $y_1 + y_2 + \cdots + y_n = 1$ and $0 \leq y_1, y_2, \ldots, y_n \leq \frac{1}{2}$, the following inequality holds:
\[ x_1 x_2 \cdots x_n \leq \gamma \left( x_1 y_1 + x_2 y_2 + \cdots + x_n y_n \right), \]
we need to maximize
\[ \lambda = \frac{x_1 x_2 \cdots x_n}{x_1 y_1 + x_2 y_2 + \cdots + x_n y_n}. \]

1. **Assume without loss of generality that $x_1 \ge x_2 \ge \cdots \ge x_n$.**

2. **To maximize $\lambda$, we need to minimize $x_1 y_1 + x_2 y_2 + \cdots + x_n y_n$.**

3. **Given the constraints $0 \leq y_i \leq \frac{1}{2}$ and $y_1 + y_2 + \cdots + y_n = 1$, the minimum value of $x_1 y_1 + x_2 y_2 + \cdots + x_n y_n$ occurs when $y_{n-1} = y_n = \frac{1}{2}$ and $y_1 = y_2 = \cdots = y_{n-2} = 0$.**

4. **Thus, the minimum value is:**
\[ x_1 y_1 + x_2 y_2 + \cdots + x_n y_n = \frac{1}{2} (x_{n-1} + x_n). \]

5. **Therefore, we have:**
\[ \lambda \leq \frac{2 x_1 x_2 \cdots x_n}{x_{n-1} + x_n}. \]

6. **Using the AM-GM inequality, we know:**
\[ x_{n-1} + x_n \geq 2 \sqrt{x_{n-1} x_n}. \]

7. **Thus:**
\[ \lambda \leq \frac{2 x_1 x_2 \cdots x_n}{2 \sqrt{x_{n-1} x_n}} = x_1 x_2 \cdots x_{n-2} \sqrt{x_{n-1} x_n}. \]

8. **Let $a$ be the arithmetic mean of $x_1, x_2, \ldots, x_{n-2}$ and $b$ be the arithmetic mean of $x_{n-1}$ and $x_n$. Clearly, $a \ge b$ and $(n-2)a + 2b = 1$.**

9. **Using the AM-GM inequality again:**
\[ x_1 x_2 \cdots x_{n-2} \sqrt{x_{n-1} x_n} \leq a^{n-2} b. \]

10. **Since $a \ge b$ and $(n-2)a + 2b = 1$, we have:**
\[ a = \frac{1 - 2b}{n-2}. \]

11. **Substituting $a$ into the inequality:**
\[ a^{n-2} b = \left( \frac{1 - 2b}{n-2} \right)^{n-2} b. \]

12. **To find the maximum value of this expression, we set $b = \frac{1}{n-1}$, which gives $a = \frac{1}{n-1}$ as well.**

13. **Thus:**
\[ a^{n-2} b = \left( \frac{1}{n-1} \right)^{n-2} \cdot \frac{1}{n-1} = \frac{1}{(n-1)^{n-1}}. \]

14. **Therefore, the least value of $\gamma$ is:**
\[ \gamma = \frac{1}{2(n-1)^{n-1}}. \]

The final answer is $\boxed{\frac{1}{2(n-1)^{n-1}}}$.