5. (12th Turkish Mathematical Olympiad) Given the set $K(n, 0)=\varnothing$. For any non-negative integers $m$, $n$, define $K(n, m+1)=\{k \mid 1 \leqslant k \leqslant n, K(k, m) \cap K(n-k, m)=\varnothing\}$, find the number of elements in the set $K(2004, 2004)$.
5. In the following discussion of $K(n, m)$, it is stipulated that $m \geqslant n$.

It is easy to see that $K(0, m)=\varnothing, n \in K(n, m)$.
First, list several items to find patterns.
$$
\begin{array}{l}
K(1, m)=\{1\}, K(2, m)=\{2\} ; \\
K(3, m)=\{1,2,3\}, K(4, m)=\{4\} ; \\
K(5, m)=\{1,4,5\}, K(6, m)=\{2,4,6\}, \\
K(7, m)=\{1 \sim 7\}, K(8, m)=\{8\} ; \\
K(9, m)=\{1,8,9\}, K(10, m)=\{2,8,10\}, \\
K(11, m)=\{1,2,3,8,9,10,11\}, \\
K(12, m)=\{4,8,12\}, \\
K(13, m)=\{1,4,5,8,9,12,13\}, \\
K(14, m)=\{2,4,6,8,10,12,14\}, \\
K(15, m)=\{1 \sim 15\}, K(16, m)=\{16\} .
\end{array}
$$

From calculations and definitions, it is known that $K(n, m)$ depends only on $n$, not on $m$.
Below, we use mathematical induction to prove two lemmas.
Lemma 1: $K(2 n, m)=\{2 j \mid j \in K(n, m)\}$.
Proof of Lemma 1: Assume the proposition holds for positive integers less than $n$.
By induction, we can prove $1 \in K(2 n-1, m), 1 \notin K(2 n, m), n \in \mathbf{N}_{+}$, so $2 j+1 \notin K(2 n, m)$.
Also, $2 j \in K(2 n, m) \Leftrightarrow K(2 j, m) \cap K(2 n-2 j, m)=\varnothing$
$\Leftrightarrow K(j, m) \cap K(n-j, m)=\varnothing$ (by induction hypothesis)
$$
\Leftrightarrow j \in K(n, m) \text {. }
$$

Lemma 2: $K\left(2^{n}+i, m\right)=\left\{2^{n}+i\right\} \cup K(i, m) \cup\left\{2^{n}+i-j \mid j \in K(i, m)\right\}$, where $1 \leqslant i<2^{n}$.
Proof of Lemma 2: Assume equation (1) holds for positive integers less than $2^{n}+i$.
For any $j \in K(i, m)$, if $j<i$, then
$K(j, m) \cap K(i-j, m)=\varnothing$
$\Rightarrow K(j, m) \cap K\left(2^{n}+i-j, m\right)=\varnothing$ (by induction hypothesis)
$\Rightarrow j \in K\left(2^{n}+i, m\right)$.
If $j=i$, by $i \in K(i, m), K(j, m) \cap\left\{2^{n}\right\}=\varnothing$, i.e., $K(i, m) \cap K\left(2^{n}, m\right)=\varnothing \Rightarrow i \in K\left(2^{n}+i, m\right)$.
Therefore, $K(i, m) \subset K\left(2^{n}+i, m\right)$.
By definition, $\left\{2^{n}+i-j \mid j \in K(i, m)\right\} \subset K\left(2^{n}+i, m\right), 2^{n}+i \in K\left(2^{n}+i, m\right)$.
In summary, the right-hand side of equation (1) $\subseteq K\left(2^{n}+i, m\right)$.
For any $j \in K\left(2^{n}+i, m\right)$, by definition, assume $j<\frac{2^{n}+i}{2}$. If $j \leqslant i$, then
$K(j, m) \cap K\left(2^{n}+i-j, m\right)=\varnothing$
$\Rightarrow K(j, m) \cap K(i-j, m)=\varnothing$ (by induction hypothesis)
$\Rightarrow j \in K(i, m)$.
Next, we prove that when $i<j<\frac{2^{n}+i}{2}$, $j \notin K\left(2^{n}+i, m\right)$.
Let $j=i+k$, at this time $k<2^{n-1}$, we only need to prove
$K(i+k, m) \cap K\left(2^{n}-k, m\right) \neq \varnothing$.
By equation (1), for any $k<2^{n}$, comparing binary, let $k=\sum_{s=1}^{r} 2^{a_{s}}$, where $0 \leqslant a_{1}<a_{2}<\cdots<a_{r}$, then $\left\{2^{\alpha} \mid s=1,2, \cdots, r\right\} \subset K(k, m)$.
If $i+k$ does not carry in binary, then
$$
2^{a_{1}} \in K(i+k, m) \cap K(k, m) \cap K\left(2^{n}-k, m\right) .
$$

If $i+k$ carries in binary, assume it carries for the last time at the $t$-th bit, by $j<\frac{2^{n}+i}{2}$, we know $t<n$, then the coefficient of $2^{t}$ in $i, k$ is 0, and the coefficient of $2^{t}$ in $i+k$ is 1. At this time, the coefficient of $2^{t}$ in $2^{n}-k$ is 1, so $2^{t} \in K(i+k, m) \cap K\left(2^{n}-k, m\right)$.
Now, we solve the original problem. By Lemmas 1 and 2, we know
$$
\begin{aligned}
|K(2004, m)| & =|K(1002, m)|=|K(501, m)|=|K(256+245, m)| \\
& =2|K(245, m)|+1=2|K(128+117, m)|+1 \\
& =4|K(117, m)|+3=4|K(64+53, m)|+3 \\
& =8|K(53, m)|+7=8|K(32+21, m)|+7 \\
& =16|K(21, m)|+15=16|K(16+5, m)|+15 \\
& =32|K(5, m)|+31=32 \times 3+31=127 .
\end{aligned}
$$

Therefore, the number of elements in the set $K(2004,2004)$ is 127.