Find all real solutions of the system
$$\begin{cases}  x_1 +x_2 +...+x_{2000} = 2000 \\ x_1^4 +x_2^4 +...+x_{2000}^4= x_1^3 +x_2^3 +...+x_{2000}^3\end{cases}$$
1. We start with the given system of equations:
   \[
   \begin{cases}
   x_1 + x_2 + \cdots + x_{2000} = 2000 \\
   x_1^4 + x_2^4 + \cdots + x_{2000}^4 = x_1^3 + x_2^3 + \cdots + x_{2000}^3
   \end{cases}
   \]

2. Define the function \( f(x) = x^4 - x^3 - x + 1 \). We can factorize \( f(x) \) as follows:
   \[
   f(x) = (x-1)(x^3 - 1) = (x-1)(x-1)(x^2 + x + 1) = (x-1)^2(x^2 + x + 1)
   \]

3. Notice that \( x^2 + x + 1 \) is always positive for all real \( x \). Therefore, \( f(x) = (x-1)^2(x^2 + x + 1) \geq 0 \) for all real \( x \), with equality if and only if \( x = 1 \).

4. From the given equations, we can write:
   \[
   \sum_{k=1}^{2000} x_k = 2000
   \]
   and
   \[
   \sum_{k=1}^{2000} x_k^4 = \sum_{k=1}^{2000} x_k^3
   \]

5. Consider the sum of \( f(x_k) \) for \( k = 1, 2, \ldots, 2000 \):
   \[
   \sum_{k=1}^{2000} f(x_k) = \sum_{k=1}^{2000} (x_k^4 - x_k^3 - x_k + 1)
   \]

6. Using the given equations, we can rewrite the sum:
   \[
   \sum_{k=1}^{2000} f(x_k) = \sum_{k=1}^{2000} x_k^4 - \sum_{k=1}^{2000} x_k^3 - \sum_{k=1}^{2000} x_k + \sum_{k=1}^{2000} 1
   \]
   \[
   = \sum_{k=1}^{2000} x_k^4 - \sum_{k=1}^{2000} x_k^3 - \sum_{k=1}^{2000} x_k + 2000
   \]
   \[
   = \sum_{k=1}^{2000} x_k^3 - \sum_{k=1}^{2000} x_k^3 - 2000 + 2000
   \]
   \[
   = 0
   \]

7. Since \( \sum_{k=1}^{2000} f(x_k) = 0 \) and \( f(x) \geq 0 \) for all real \( x \) with equality only when \( x = 1 \), it follows that \( f(x_k) = 0 \) for all \( k \). Therefore, \( x_k = 1 \) for all \( k \).

8. Thus, the only solution to the system is \( x_1 = x_2 = \cdots = x_{2000} = 1 \).

The final answer is \( \boxed{ x_1 = x_2 = \cdots = x_{2000} = 1 } \).