[Example 5.6.2] The number 3 can be represented as the sum of one or more positive integers in four ways, such as $3, 1+2, 2+1, 1+1+1$. How many ways can 2004 be represented as the sum of one or more positive integers?
We study the partition of 3, which can be considered as follows:
Arrange 3 ones in a horizontal row:
$$
\begin{array}{lll}
1 & 1 & 1
\end{array}
$$

We can add plus signs or not add them in the gaps between them, for example, we can write:
$$
(1,1,1) \quad(1,1+1),(1+1,1),(1+1+1) \text {. }
$$

Each parenthesis corresponds to $3,1+2,2+1$, and $1+1+1$.
From this, we know that for 2004, we can write 2004 ones in a horizontal row:
$$
\begin{array}{lllllll}
1 & 1 & 1 & 1 & \cdots & 1 & 1
\end{array}
$$

There are 2003 gaps, and for each gap, we can add a plus sign (“+”) or not add one. Thus, for each gap, there are two possibilities, resulting in a total of $2^{2003}$ possibilities. That is, 2004 can be expressed as the sum of one or several positive integers in $2^{2003}$ different ways.

We can also understand it this way: for the 2004 ones in (1) and the 2003 gaps between them, we can add 0 “+” signs, 1 “+” sign, 2 “+” signs, ..., up to 2003 “+” signs. There are
$$
C_{2003}^{0}+C_{2003}^{1}+\cdots+C_{2003}^{2003}=2^{2003}
$$

different ways to add “+” signs, meaning there are $2^{2003}$ solutions to this problem.