Example 6 Let $X=\{1,2, \cdots, 2001\}$, find the smallest positive integer $m$, such that for any $m$-element subset $W$ of $X$, there exist $u, v \in W$ ($u$ and $v$ can be the same), such that $u+v$ is a power of 2. (2001 China Mathematical Olympiad Problem)
Analysis and Solution: For the convenience of describing the problem, if $u+v$ is a power of 2, then $u, v$ is called a pair. We consider the problem from the opposite side. If the subset $W$ of $X$ does not contain any pairs, then what is the maximum number of elements in $W$? Clearly, if we can divide $X$ into several blocks such that any two numbers in each block form a pair, then $W$ can only contain one element from each block. Therefore, let $A_{i}=\{1024-i, 1024+i\}(i=1,2, \cdots, 977)$, $B_{j}=\{32-j, 32+j\}(j=1,2, \cdots, 14)$, $C=\{15,17\}$, $D_{k}=\{8-k, 8+k\}(k=1,2, \cdots, 6)$, $E=\{1,8,16,32,1024\}$. Assuming $W$ does not contain any pairs, then $W$ cannot contain any elements from $E$, and can contain at most one element from each $A_{i}$, $B_{j}$, $D_{k}$, and $C$. Thus, $|W| \leqslant 977+14+6+1=998$. This indicates that when $|W| \geqslant 999$, $W$ must contain a pair, meaning that $m=999$ meets the condition. 

Furthermore, if $|W|=998$ and $W$ does not contain any pairs, then $W$ must contain exactly one element from each $A_{i}$, $B_{j}$, $D_{k}$, and $C$. Let $W=\{1025,1026, \cdots, 2001\} \cup\{33,34, \cdots, 46\} \cup\{17\} \cup\{9,10,11,12,13,14\}$. It is easy to verify that $W$ does not contain any pairs. Therefore, when $m<999$, taking an $m$-element subset of $W$, the subset will not contain any pairs.

In summary, the minimum value of $m$ is 999.