In a box, there are 10 red cards and 10 blue cards, each color containing cards labeled with numbers $1, 3, 3^{2}, \cdots, 3^{9}$, one for each number. The total sum of the labels on the cards of both colors is denoted as $S$. For a given positive integer $n$, if it is possible to select some cards from the box such that the sum of their labels is exactly $n$, this is called a "scheme for $n$", and the number of different schemes for $n$ is denoted as $f(n)$. Find $P=$ $\sum_{n=1}^{1000} f(n)$.
Let the maximum sum of the labels of two-color cards marked as $1,3,3^{2}, \cdots, 3^{k}$ be denoted as $S_{k}$. Then,
$$
S_{k}=2 \sum_{n=0}^{k} 3^{n}=3^{k+1}-1<3^{k+1} \text {. }
$$

In the sequence $1,3,3^{2}, \cdots, 3^{k}$, the sum of any subset of these numbers is not equal to $3^{m}$. Therefore, the number of ways to pick cards labeled with $3^{k}$ is $\mathrm{C}_{2}^{0}+\mathrm{C}_{2}^{1}+\mathrm{C}_{2}^{2}=4$, and each picking method corresponds to a unique scheme. Thus, the total number of ways to pick cards labeled with $1,3,3^{2}, \cdots, 3^{k-1}$ is $4^{k}-1$ (excluding the case where no card is picked).
Notice that, $1000=3^{6}+3^{5}+3^{3}+1$.
By repeatedly applying the above conclusion, we get
$$
\begin{aligned}
P= & \sum_{n=1}^{1000} f(n) \\
= & \sum_{n=1}^{3^{6-1}} f(n)+f\left(3^{6}\right)+\sum_{n=3^{6}+1}^{3^{6}+3^{5}-1} f(n)+ \\
& f\left(3^{6}+3^{5}\right)+\sum_{n=3^{6}+3^{5}+1}^{3^{6}+3^{5}+3^{3}-1} f(n)+ \\
& f\left(3^{6}+3^{5}+3^{3}\right)+f\left(3^{6}+3^{5}+3^{3}+1\right) \\
= & 4^{6}-1+f\left(3^{6}\right)+2\left(4^{5}-1\right)+f\left(3^{6}+3^{5}\right)+ \\
& 4\left(4^{3}-1\right)+f\left(3^{6}+3^{5}+3^{3}\right)+ \\
& f\left(3^{6}+3^{5}+3^{3}+1\right) \\
= & 4^{6}-1+2+2\left(4^{5}-1\right)+2^{2}+ \\
& 4\left(4^{3}-1\right)+2^{3}+2^{4} \\
= & 6423 .
\end{aligned}
$$
(Zheng Kai, Changshu High School, Jiangsu Province, 215500)