Pablo copied from the blackboard the problem:
[list]Consider all the sequences of $2004$ real numbers $(x_0,x_1,x_2,\dots, x_{2003})$ such that: $x_0=1, 0\le x_1\le 2x_0,0\le x_2\le 2x_1\ldots ,0\le x_{2003}\le 2x_{2002}$. From all these sequences, determine the sequence which minimizes $S=\cdots$[/list]

As Pablo was copying the expression, it was erased from the board. The only thing that he could remember was that $S$ was of the form $S=\pm x_1\pm x_2\pm\cdots\pm x_{2002}+x_{2003}$. Show that, even when Pablo does not have the complete statement, he can determine the solution of the problem.
To solve this problem, we need to determine the sequence \((x_0, x_1, x_2, \ldots, x_{2003})\) that minimizes \(S = \pm x_1 \pm x_2 \pm \cdots \pm x_{2002} + x_{2003}\) under the given constraints. We will use induction to show that the sequence \(x_i = 2^i\) for \(i = 0, 1, 2, \ldots, 2003\) maximizes \(S\).

1. **Base Case:**
   For \(n = 1\), the sequence is \((x_0, x_1)\) with \(x_0 = 1\) and \(0 \leq x_1 \leq 2x_0 = 2\). The expression for \(S\) is \(S = \pm x_1 + x_2\). Since \(x_2\) is not defined for \(n = 1\), we consider only \(x_1\). The maximum value of \(x_1\) is \(2\), so \(S\) is maximized when \(x_1 = 2\).

2. **Inductive Step:**
   Assume that for \(n = k\), the sequence \((x_0, x_1, x_2, \ldots, x_{k-1})\) with \(x_i = 2^i\) maximizes \(S = \pm x_1 \pm x_2 \pm \cdots \pm x_{k-1} + x_k\).

   For \(n = k+1\), consider the sequence \((x_0, x_1, x_2, \ldots, x_k)\) with \(x_0 = 1\) and \(0 \leq x_i \leq 2x_{i-1}\) for \(i = 1, 2, \ldots, k\). We need to show that \(x_i = 2^i\) for \(i = 0, 1, 2, \ldots, k\) maximizes \(S = \pm x_1 \pm x_2 \pm \cdots \pm x_k + x_{k+1}\).

   - If the sign before \(x_{k-1}\) is positive, then \(S - x_k\) is maximized by the inductive hypothesis, so \(x_k\) will be maximized if \(x_i = 2^i\).
   - If the sign before \(x_{k-1}\) is negative, then \(-x_{k-1} + x_k \leq -x_{k-1} + 2x_{k-1} = x_{k-1}\), with equality if \(x_k = 2x_{k-1}\). Thus, \(S \leq \pm x_1 \pm x_2 \pm \cdots \pm x_{k-2} + x_{k-1}\), which is maximized if \(x_i = 2^i\).

   Therefore, \(S\) is maximized if \(x_i = 2^i\) for \(i = 0, 1, 2, \ldots, k\) and \(x_{k+1} = 2x_k\).

By induction, the sequence \((x_0, x_1, x_2, \ldots, x_{2003})\) that maximizes \(S\) is given by \(x_i = 2^i\) for \(i = 0, 1, 2, \ldots, 2003\).

The final answer is \( \boxed{ x_i = 2^i } \) for \(i = 0, 1, 2, \ldots, 2003\).