Find  the largest real number $c$ such that $$\sum_{i=1}^{n}\sum_{j=1}^{n}(n-|i-j|)x_ix_j \geq c\sum_{j=1}^{n}x^2_i$$
for  any positive integer $n $ and  any real numbers $x_1,x_2,\dots,x_n.$
1. We start with the given inequality:
   \[
   \sum_{i=1}^{n}\sum_{j=1}^{n}(n-|i-j|)x_ix_j \geq c\sum_{j=1}^{n}x^2_i
   \]
   We need to find the largest real number \( c \) such that this inequality holds for any positive integer \( n \) and any real numbers \( x_1, x_2, \ldots, x_n \).

2. Let \( s = \sum_{i=1}^{n} x_i \). We introduce the notation \( y_k = \frac{1}{2} \left( -\sum_{l=1}^{k} x_l + \sum_{l=k+1}^{n} x_l \right) \) with \( y_0 = -y_n = \frac{s}{2} \).

3. We rewrite the left-hand side of the inequality:
   \[
   \sum_{i,j}(n-|i-j|)x_ix_j = s^2 + \sum_{k=1}^{n-1} \left( \left( \sum_{l=1}^{k} x_l \right)^2 + \left( \sum_{l=k+1}^{n} x_l \right)^2 \right)
   \]

4. Using the definition of \( y_k \), we can express the sum as:
   \[
   s^2 + \sum_{k=1}^{n-1} \left( 2y_k^2 + \frac{s^2}{2} \right) = \frac{n-1}{2}s^2 + \sum_{k=1}^{n} (y_{k-1}^2 + y_k^2)
   \]

5. We know that:
   \[
   \sum_{k=1}^{n} (y_{k-1}^2 + y_k^2) \geq 0 + \frac{1}{2} \sum_{k=1}^{n} (y_{k-1} - y_k)^2 = \frac{1}{2} \sum_{k=1}^{n} x_k^2
   \]

6. Therefore, we have:
   \[
   \sum_{i,j}(n-|i-j|)x_ix_j \geq \frac{1}{2} \sum_{k=1}^{n} x_k^2
   \]

7. This implies that the largest value of \( c \) is \( \frac{1}{2} \).

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