Let $0=x_0<x_1<\cdots<x_n=1$ .Find the largest real number$ C$ such  that for any positive integer $ n $ ,  we have $$\sum_{k=1}^n  x^2_k (x_k - x_{k-1})>C$$
1. **Solution 1:**
   - For each integer \( n > 0 \), define \( c_n \) as the best constant satisfying 
     \[
     \sum_{k=1}^n x_k^2(x_k - x_{k-1}) \ge c_n x_n^3
     \]
   - For \( n+1 \), the inequality 
     \[
     c_n x_n^3 - x_n x_{n+1}^2 + (1 - c_{n+1}) x_{n+1}^3 \ge 0
     \]
     needs to hold for any real \( x_n < x_{n+1} \).
   - We have \( c_1 = 1 \). By mathematical induction, we get 
     \[
     \frac{1}{3} < c_{n+1} = 1 - 2\sqrt{\frac{1}{27c_n}} < c_n
     \]
   - The \( c \) we need is 
     \[
     \lim_{n \to \infty} c_n = \frac{1}{3}
     \]

2. **Solution 2:**
   - Take \( x_k = \frac{k}{n}, k = 1, 2, \ldots, n \). We get 
     \[
     \sum_{k=1}^n x_k^2(x_k - x_{k-1}) = \sum_{k=1}^n \left( \frac{k}{n} \right)^2 \left( \frac{k}{n} - \frac{k-1}{n} \right) = \sum_{k=1}^n \left( \frac{k}{n} \right)^2 \left( \frac{1}{n} \right)
     \]
     \[
     = \frac{1}{n^3} \sum_{k=1}^n k^2 = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2}
     \]
   - This holds for every \( n \), thus \( c \le \frac{1}{3} \).
   - Now we prove \( c = \frac{1}{3} \) is valid. By AM-GM, 
     \[
     x_k^3 + x_k^3 + x_{k-1}^3 - 3x_{k-1}x_k^2 \ge 0, \quad (k = 1, 2, \ldots, n)
     \]
     \[
     \sum_{k=1}^n x_k^2(x_k - x_{k-1}) - \frac{1}{3} x_n^3 = \frac{2}{3} x_1^3 + \frac{1}{3} \sum_{k=2}^n (x_k^3 + x_k^3 + x_{k-1}^3 - 3x_{k-1}x_k^2) \ge 0
     \]

The final answer is \( \boxed{\frac{1}{3}} \)