Example 2 Find $\sum_{k=0}^{1003} \mathrm{C}_{2006-k}^{k}$.
Explanation: Transform this problem into:
Climbing a 2006-step staircase, one can take 1 step at a time or 2 steps at a time. How many different ways are there to climb it?

On one hand, classify by the number of times taking 2 steps at a time, there are 0, $1, \cdots, 1003$ for a total of 1004 cases (as shown in Table 1).
Table 1
\begin{tabular}{|c|c|c|}
\hline 2006 steps to finish & \begin{tabular}{l}
0 times taking 2 steps, each step \\
taking 1 step
\end{tabular} & $\mathrm{C}_{2000}^{0}$ ways \\
\hline 2005 steps to finish & \begin{tabular}{l}
1 time taking 2 steps, 2004 steps \\
taking 1 step
\end{tabular} & $\mathrm{C}_{2 \cos \text { ways }}^{1}$ \\
\hline ...... & ...... & ...... \\
\hline \begin{tabular}{l}
$2006-k$ steps to \\
finish
\end{tabular} & \begin{tabular}{l}
$k$ times taking 2 steps, $2006-$ \\
$2 k$ steps taking 1 step
\end{tabular} & $\mathrm{C}_{200-k}^{\mathrm{k}}$ ways \\
\hline ...... & ...... & ...... \\
\hline 1003 steps to finish & \begin{tabular}{l} 
always taking 2 steps, 0 times \\
taking 1 step
\end{tabular} & $\mathrm{C}_{\infty \infty 0 \infty}^{10 \infty}$ \\
\hline
\end{tabular}

On the other hand, let the number of ways to climb $n$ steps be $a_{n}$. If the first step is 1 step, then the remaining $n-1$ steps have $a_{n-1}$ ways; if the first step is 2 steps, then the remaining $n-2$ steps have $a_{n-2}$ ways. Thus, we have
$$
a_{n}=a_{n-1}+a_{n-2},
$$

where, $a_{1}=1, a_{2}=2$.
Therefore, the sum of combinations (using the technique of constructing and calculating twice) is transformed into finding the general term of the Fibonacci sequence, giving
$$
a_{2006}=\sum_{k=0}^{1003} \mathrm{C}_{2000-k}^{k} .
$$