2・71 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 $u \in U$, $(s, u)>1$.
[Solution] Notice 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 now 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 $S=\left\{s_{1}, s_{2}, s_{3}, s_{4}\right\}, s_{1}1000,
$$

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

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 mentioned above, and by the pigeonhole principle, at least 4 triples belong to the same group among these 13 groups. Writing these 4 triples as a $3 \times 4$ table, the 3 rows of numbers are $S, T$, and $U$, respectively, which meet the requirements of the problem.
In summary, the smallest natural number $n=982$.