4. Let $A=\{1,2,3, \cdots, 1997\}$, for any 999-element subset $X$ of $A$, if there exist $x, y \in X$, such that $x<y$ and $x \mid y$, then $X$ is called a good set. Find the largest natural number $a(a \in A)$, such that any 999-element subset containing $a$ is a good set.
(《Mathematics in Middle School》1999 Issue 1 Olympiad Problems)
4. The desired $\max a=665$. First, prove $a \leqslant 665$. Clearly, the 999-element subset $X_{0}=\{999,1000,1001, \cdots, 1997 \mid$ does not contain $x, y \in X_{0}$ such that $x \cdot 1997$, which is greater than the largest element in $X_{0}$, thus it holds. Therefore, $a$ cannot be any of the numbers $999,1000,1001, \cdots, 1997$.
Next, prove $a \neq 666,667,668, \cdots, 997,998$. Construct the set:
$B_{i}=\{666+i\} \cup X_{0} \backslash\{2(666+i)\}, i=0,1, \cdots, 332$.
For $B_{i}$, $(666+i) \cdot 3 \geqslant 1998$, and $(666+i) \cdot 2 \notin B_{i}$, so no other multiples of $666+i$ are in $B_{i}$. It has already been proven that any non-multiple of elements in $X_{0}$ are not in $X_{0}$, and $666+i<999(i=0,1,2, \cdots, 332)$, so no multiple of any element in $X_{0}$ can be $666+i(i=0,1, \cdots, 332)$. Thus, the multiples (excluding itself) of any other elements in $B_{i}$ are not in $B_{i}$, ensuring that there are no two elements in $B_{i}$ such that $x<y, x \mid y$. Since $B_{i}$ (for $i=0,1, \cdots, 332$) includes 666, 667, ..., 998, it follows that $a \neq 666,667, \cdots, 998$, thus $a \leqslant 665$.

Next, prove that 665 is achievable. Assume there exists a 999-element subset $X$ containing 665 such that no $x, y \in X, x<y$ satisfies $x \mid y$. Then $665 \cdot 2, 665 \cdot 3 \notin X$. Construct 997 drawers that include all elements of $A$ except $665, 665 \cdot 2, 665 \cdot 3$, with each element appearing once: $K_{1}=\left|1,1 \cdot 2,1 \cdot 2^{2}, \cdots, 1 \cdot 2^{10}\right|, K_{2}=\left|3,3 \cdot 2,3 \cdot 2^{2}, \cdots, 3 \cdot 2^{9}\right|, K_{3}=\mid 5,5 \cdot 2$, $\left.5 \cdot 2^{2}, \cdots, 5 \cdot 2^{8}\right\}, K_{4}=\left\{7,7 \cdot 2,7 \cdot 2^{2}, \cdots, 7 \cdot 2^{8}\right\}, \cdots, K_{332}=\{663,663 \cdot 2,663 \cdot 3\}, K_{334}=\{667,667 \cdot$ $2 \mid, K_{335}=\{669,669 \cdot 2\}, \cdots, K_{995}=\{1991\}, K_{996}=\{1993\}, K_{997}=\{1997\}$. Thus, the other 998 elements of $X$ (excluding 665) are placed into these 997 drawers, meaning two elements must be in the same drawer, and elements in the same drawer are multiples of each other, leading to a contradiction. Proof complete.