Example 12 Let $X=\{1,2, \cdots, 2001\}$. Find the smallest positive integer $m$, such that: for any $m$-element subset $W$ of $X$, there exist $u, v \in W$ (where $u$ and $v$ can be the same), such that $u+v$ is a power of 2.

保留源文本的换行和格式，翻译结果如下：

Example 12 Let $X=\{1,2, \cdots, 2001\}$. Find the smallest positive integer $m$, such that: for any $m$-element subset $W$ of $X$, there exist $u, v \in W$ (where $u$ and $v$ can be the same), such that $u+v$ is a power of 2.
When $u$ and $v$ are $2^{r} + a$ and $2^{r} - a$ respectively, $u + v = 2 \times 2^{r} = 2^{r+1}$ is a power of 2. Thus, $X$ can be divided into the following 5 subsets:
$$
\begin{array}{l}
2001 = 2^{10} + 977 \geqslant x \geqslant 2^{10} - 977 = 47, \\
46 = 2^{5} + 14 \geqslant x \geqslant 2^{5} - 14 = 18, \\
17 = 2^{4} + 1 \geqslant x \geqslant 2^{4} - 1 = 15, \\
14 = 2^{3} + 6 \geqslant x \geqslant 2^{3} - 6 = 2, \\
x = 1. \\
\text{Let } Y = \{2001, 2000, \cdots, 1025\} \cup \{46, 45, \cdots, 33\} \cup \{17\} \cup \{14, 13, \cdots, 9\}.
\end{array}
$$

Thus, $|Y| = 998$, and for any $u, v \in Y$, $u + v$ is not a power of 2.
The proof is as follows:
When $u, v \in Y$, assume $u \geqslant v$ and $2^{r} < u \leqslant 2^{r} + a < 2^{r+1}$, where, when $r$ takes the values 10, 5, 4, 3, the corresponding $a$ values are
$$
977, 14, 1, 6.
$$
(1) If $2^{r} < v \leqslant u$, then $2^{r+1} < u + v < 2^{r+2}$, so $u + v$ cannot be a power of 2.
(2) If $1 \leqslant v < 2^{r}$, then when $2^{r} < u \leqslant 2^{r} + a (1 \leqslant a < 2^{r})$, $1 \leqslant v < 2^{r} - a$. Thus, $2^{r} < u + v < 2^{r+1}$, so $u + v$ cannot be a power of 2.
Therefore, the sum of any two numbers in the subset $Y$ is not a power of 2.
Hence, the smallest positive integer $m \geqslant 999$.
Divide $X$ into 999 mutually disjoint subsets:
$$
\begin{array}{l}
A_{i} = \{1024 - i, 1024 + i\} (i = 1, 2, \cdots, 977), \\
B_{j} = \{32 - j, 32 + j\} (j = 1, 2, \cdots, 14), \\
C = \{15, 17\}, \\
D_{k} = \{8 - k, 8 + k\} (k = 1, 2, \cdots, 6), \\
E = \{1, 8, 16, 32, 1024\}.
\end{array}
$$

For any 999-element subset $W$ of $X$, if $W \cap E \neq \varnothing$, then the double of any element taken from it is a power of 2; if $W \cap E = \varnothing$, then the 999 elements of $W$ belong to the preceding 998 2-element subsets. By the pigeonhole principle, $W$ must contain different $u, v$ that belong to the same subset. Clearly, $u + v$ is a power of 2. In summary, the smallest positive integer $m = 999$.