11. (1996 China National Team Selection Contest for IMO) Let $M=\{2,3,4, \cdots, 1000\}$. Find the smallest natural number $n$, such that in any $n$-element subset of $M$, there exist 3 pairwise disjoint 4-element subsets $S, T, U$, satisfying the following conditions:
（1）For any two elements in $S$, the larger number is a multiple of the smaller number, and the same property holds for $T$ and $U$:
（2）For any $s \in S$ and $t \in T$, $(s, t)=1$;
（3）For any $s \in S$ and $u \in U$, $(s, u)>1$.
11. Noting that $999=37 \times 27$, we let 
$$
\left\{\begin{array}{l}
A=\{3,5, \cdots, 37\}, \\
B=M-A .
\end{array}\right.
$$

Clearly, $|A|=18,|B|=981$.
We will prove that the subset $B$ of $M$ cannot simultaneously satisfy conditions (1) to (3).
We use proof by contradiction.
If $B$ contains 3 mutually disjoint 4-element subsets $S, T, U$.

Let
$$
\begin{array}{l}
S=\left\{s_{1}, s_{2}, s_{3}, s_{4}\right\}, s_{1}<s_{2}<s_{3}<s_{4}, \\
T=\left\{t_{1}, t_{2}, t_{3}, t_{4}\right\}, t_{1}<t_{2}<t_{3}<t_{4}, \\
U=\left\{u_{1}, u_{2}, u_{3}, u_{4}\right\}, u_{1}<u_{2}<u_{3}<u_{4}.
\end{array}
$$

Since $S, T, U$ are mutually disjoint, we have
$$
s_{1}+t_{1}+u_{1} \geq 1+2+3=6, \\
s_{2}+t_{2}+u_{2} \geq 4+5+6=15, \\
s_{3}+t_{3}+u_{3} \geq 7+8+9=24, \\
s_{4}+t_{4}+u_{4} \geq 10+11+12=33.
$$

Thus,
$$
(s_{1}+s_{2}+s_{3}+s_{4})+(t_{1}+t_{2}+t_{3}+t_{4})+(u_{1}+u_{2}+u_{3}+u_{4}) \geq 6+15+24+33=78.
$$

However, since $B$ contains 981 elements, the sum of the elements in $B$ is at most
$$
1+2+\cdots+981 = \frac{981 \times 982}{2} = 481531.
$$

Since $S, T, U$ are 4-element subsets of $B$, the sum of the elements in $S, T, U$ is at most
$$
481531 - (3+5+\cdots+37) = 481531 - 342 = 481189.
$$

But,
$$
78 \times 3 = 234,
$$

which is less than 481189. This is a contradiction.
This shows that the smallest natural number $n \geq 982$.
On the other hand, let
$$
\begin{array}{l}
\left\{\begin{array}{l}
S_{1}=\{3,9,27,81,243,729\}, \\
T_{1}=\{2,4,8,16,32,64\}, \\
U_{1}=\{6,12,24,48,96,192\};
\end{array}\right. \\
\left\{\begin{array}{l}
S_{2}=\{5,15,45,135,405\}, \\
T_{2}=\{41,82,164,328,656\}, \\
U_{2}=\{10,20,40,80,160\};
\end{array}\right. \\
\left\{\begin{array}{l}
S_{3}=\{7,21,63,189,567\}, \\
T_{3}=\{43,86,172,344,688\}, \\
U_{3}=\{14,28,56,112,224\};
\end{array}\right. \\
\left\{\begin{array}{l}
S_{4}=\{11,33,99,297,891\}, \\
T_{4}=\{47,94,188,376,752\}, \\
U_{4}=\{22,44,88,176,352\};
\end{array}\right. \\
\left\{\begin{array}{l}
S_{5}=\{13,39,117,351\}, \\
T_{5}=\{53,106,212,424\}, \\
U_{5}=\{26,52,104,208\};
\end{array}\right. \\
\left\{\begin{array}{l}
S_{6}=\{17,51,153,459\}, \\
T_{6}=\{59,118,236,472\}, \\
U_{6}=\{34,68,136,272\};
\end{array}\right. \\
\left\{\begin{array}{l}
S_{7}=\{19,57,171,513\}, \\
T_{7}=\{61,122,244,488\}, \\
U_{7}=\{38,76,152,304\};
\end{array}\right. \\
\left\{\begin{array}{l}
S_{8}=\{23,69,207,621\}, \\
T_{8}=\{67,134,268,536\}, \\
U_{8}=\{46,92,184,368\};
\end{array}\right. \\
\left\{\begin{array}{l}
S_{9}=\{25,75,225,675\}, \\
T_{9}=\{71,142,284,568\}, \\
U_{9}=\{50,100,200,400\};
\end{array}\right. \\
\left\{\begin{array}{l}
S_{10}=\{29,87,261,783\}, \\
T_{10}=\{73,146,292,584\}, \\
U_{10}=\{58,116,232,464\};
\end{array}\right. \\
\left\{\begin{array}{l}
S_{11}=\{31,93,279,837\}, \\
T_{11}=\{79,158,316,632\}, \\
U_{11}=\{62,124,248,496\};
\end{array}\right. \\
\left\{\begin{array}{l}
S_{12}=\{35,105,315,945\}, \\
T_{12}=\{83,166,332,664\}, \\
U_{12}=\{70,140,280,560\};
\end{array}\right. \\
\left\{\begin{array}{l}
S_{13}=\{37,111,333,999\}, \\
T_{13}=\{89,178,356,712\}, \\
U_{13}=\{74,148,296,592\}.
\end{array}\right.
\end{array}
$$

We can form 57 triples by combining the 3 numbers with the same index from $S_{i}, T_{i}, U_{i}$. For any 982-element subset $B$ of $M$, only 17 numbers in $M$ are not in $B$, so at least 40 of the 57 triples are contained in $B$. These 40 triples belong to the 13 groups, and by the pigeonhole principle, at least 4 triples belong to the same group. Writing these 4 triples as a $3 \times 4$ table, the 3 rows are $S, T, U$, which satisfy the requirements of the problem.
In conclusion, the smallest natural number $n=982$.