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.
（Supplied by Zhang Zhusheng）
To divide $X$ into the following 5 subsets for examination:
$$
\begin{array}{l}
2001=1024+977 \geqslant x \geqslant 1024-977=47, \\
46=32+14 \geqslant x \geqslant 32-14=18, \\
17=16+1 \geqslant x \geqslant 16-1=15, \\
14=8+6 \geqslant x \geqslant 8-6=2, \\
x=1 .
\end{array}
$$

To construct an example that does not satisfy the requirements of the problem and yet contains the maximum number of elements, this subset must not contain any power of 2, and in each pair $\left\{2^{r}+a, 2^{r}-a\right\}$, only one can be included in the set. Let
$$
Y=\{2001,2000, \cdots, 1025\} \cup\{46,45, \cdots, 33\}
$$
$\cup\{17\} \cup\{14,13, \cdots, 9\} \cup\{1\}$,
then $|Y|=998$ and for any $u, v \in Y, u+v$ is not a power of 2.

In fact, when $u, v \in Y$, assume without loss of generality that $u \geqslant v$ and there is $2^{r}<u \leqslant 2^{r}+a<2^{r+1}$, where when $r$ takes the values $10,5,4,3$ respectively, the corresponding $a$ values are $977,14,1,6$.
(1) If $2^{r}<v \leqslant u$, then
$$
2^{r+1}<u+v<2^{r+2} \text {. }
$$
$u+v$ cannot be a power of 2.
(2) If $1 \leqslant v<2^{r}$, then when $2^{r}<u \leqslant 2^{r}+a, 1 \leqslant a<2^{r}$, we have $1 \leqslant v<2^{r}-a$. Thus
$$
2^{r}<u+v<2^{r+1}
$$

This shows that $u+v$ 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 $m \geqslant 999$.
$$
\begin{array}{l}
A_{i}:=\{1524-i, 1024+i\} ; i=1,2, \cdots, 977, \\
B=\{32-j, 32+j\}, j=1,2, \cdots, 14, \\
C=\{15,17\}, \\
D_{k}=\{8-k, 8+k\}, k=1,2, \cdots, 6, \\
E=\{1,8,16,32,1024\} .
\end{array}
$$

For any 999-element subset $W$ of $S$, if $W \cap E \neq \varnothing$, then taking any element from it, its double is a power of 2; if $W \cap E=\varnothing$, then the 999 elements of $W$ belong to the preceding 998 two-element subsets. By the pigeonhole principle, there must be different $u$ and $v$ in $W$ belonging to the same subset. Clearly, $u+v$ is a power of 2.
In summary, the smallest positive integer $m=999$.