Let $n\geq 2$ is positive integer. Find the best constant $C(n)$ such that \[\sum_{i=1}^{n}x_{i}\geq C(n)\sum_{1\leq j<i\leq n}(2x_{i}x_{j}+\sqrt{x_{i}x_{j}})\] is true for all real numbers $x_{i}\in(0,1),i=1,...,n$ for which $(1-x_{i})(1-x_{j})\geq\frac{1}{4},1\leq j<i \leq n.$
To find the best constant \( C(n) \) such that 
\[
\sum_{i=1}^{n} x_i \geq C(n) \sum_{1 \leq j < i \leq n} \left(2x_i x_j + \sqrt{x_i x_j}\right)
\]
is true for all real numbers \( x_i \in (0,1) \) for which \((1-x_i)(1-x_j) \geq \frac{1}{4}\) for \(1 \leq j < i \leq n\), we proceed as follows:

1. **Condition Analysis**:
   The condition \((1-x_i)(1-x_j) \geq \frac{1}{4}\) implies that for all \(i\) and \(j\),
   \[
   (1-x_i)(1-x_j) \geq \frac{1}{4}.
   \]
   This can be rewritten as:
   \[
   1 - x_i - x_j + x_i x_j \geq \frac{1}{4}.
   \]
   Rearranging terms, we get:
   \[
   x_i x_j - x_i - x_j \geq -\frac{3}{4}.
   \]
   Adding 1 to both sides:
   \[
   x_i x_j - x_i - x_j + 1 \geq \frac{1}{4}.
   \]
   This inequality holds for \(x_i, x_j \in (0,1)\).

2. **Symmetry and Equal Variables**:
   Consider the case where all \(x_i\) are equal, i.e., \(x_i = x\) for all \(i\). This simplifies the problem and helps in finding the best constant \(C(n)\).

3. **Simplified Expression**:
   If \(x_i = x\) for all \(i\), then:
   \[
   \sum_{i=1}^{n} x_i = nx,
   \]
   and
   \[
   \sum_{1 \leq j < i \leq n} \left(2x_i x_j + \sqrt{x_i x_j}\right) = \sum_{1 \leq j < i \leq n} \left(2x^2 + x\right).
   \]
   The number of terms in the sum \(\sum_{1 \leq j < i \leq n}\) is \(\binom{n}{2} = \frac{n(n-1)}{2}\). Therefore,
   \[
   \sum_{1 \leq j < i \leq n} \left(2x^2 + x\right) = \frac{n(n-1)}{2} \left(2x^2 + x\right).
   \]

4. **Inequality Simplification**:
   The inequality becomes:
   \[
   nx \geq C(n) \cdot \frac{n(n-1)}{2} \left(2x^2 + x\right).
   \]
   Simplifying, we get:
   \[
   2nx \geq C(n) n(n-1) (2x^2 + x).
   \]
   Dividing both sides by \(n\):
   \[
   2x \geq C(n) (n-1) (2x^2 + x).
   \]
   Dividing both sides by \(x\) (assuming \(x \neq 0\)):
   \[
   2 \geq C(n) (n-1) (2x + 1).
   \]

5. **Finding \(C(n)\)**:
   To find the best constant \(C(n)\), we need to consider the maximum value of \(2x + 1\) for \(x \in (0,1)\). The maximum value of \(2x + 1\) is 3 when \(x = 1\). Therefore,
   \[
   2 \geq 3C(n) (n-1).
   \]
   Solving for \(C(n)\), we get:
   \[
   C(n) \leq \frac{2}{3(n-1)}.
   \]
   The best constant \(C(n)\) is thus:
   \[
   C(n) = \frac{2}{3(n-1)}.
   \]

The final answer is \(\boxed{\frac{2}{3(n-1)}}\).