Three, 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 three 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: Note that $999=37 \times 27$, let $A=\{3,5, \cdots, 37\}$, $B=M-A$, then $,|A|=18,|B|=981$.

Below we prove that the subset $B$ of $M$ cannot simultaneously satisfy conditions (1) $\sim$ (3). If not, let $S=\left\{s_{1}, s_{2} ; s_{3}, s_{4}\right\}, T=\left\{t_{1}, t_{2}, t_{3}, t_{4}\right\}$ and there is $s_{1}1000$. Contradiction. This indicates that the smallest natural number $n \geqslant 982$.
On the other hand, let
$\left\{\begin{array}{l}S_{1}=\{3,9,27,81,243,729\}, \\ T_{1}=\{2,4,8,16,32,64\}, \\ U_{1}=\{6,12,24,48,96,192\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{2}=\{5,15,45,135,405\}, \\ T_{2}=\{41,82,164,328,656\}, \\ U_{2}=\{10,20,40,80,160\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{3}=\{7,21,63,189,567\}, \\ T_{3}=\{43,86,172,344,688\}, \\ U_{3}=\{14,28,56,112,224\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{4}=\{11,33,99,297,891\}, \\ T_{4}=\{47,94,188,376,752\}, \\ U_{4}=\{22,44,88,176,352\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{5}=\{13,39,117,351\}, \\ T_{5}=\{53,106,212,424\}, \\ U_{5}=\{26,52,104,208\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{6}=\{17,51,153,459\}, \\ T_{6}=\{59,118,236,472\}, \\ U_{6}=\{34,68,136,272\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{7}=\{19,57,171,513\}, \\ T_{7}=\{61,122,244,488\}, \\ U_{7}=\{38,76,152,304\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{8}=\{23,69,207,621\}, \\ T_{8}=\{67,134,268,536\}, \\ U_{8}=\{46,92,184,368\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{9}=\{25,75,225,675\}, \\ T_{9}=\{71,142,284,568\}, \\ U_{9}=\{50,100,200,400\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{10}=\{29,87,261,783\}, \\ T_{10}=\{73,146,292,584\}, \\ U_{10}=\{58,116,232,464\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{11}=\{31,93,279,837\}, \\ T_{11}=\{79,158,316,632\}, \\ U_{11}=\{62,124,248,496\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{12}=\{35,105,315,945\}, \\ T_{12}=\{83,166,332,664\}, \\ \left.U_{12}=\mid 70,140,280,560\right\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{13}=\{37,111,333,999\}, \\ T_{13}=\{89,178,356,712\}, \\ U_{13}=\{74,148,296,592\}\end{array}\right.$
Let $S_{i}, T_{i}, U_{i}$ be the 3-tuples formed by the 3 numbers in each group, a total of 57 such 3-tuples can be obtained. For any 982-element subset $B$ of $M$, only 17 numbers in $M$ are not in $B$, so at least 40 of the 57 3-tuples are contained in $B$. These 40 3-tuples belong to the 13 groups mentioned above. By the pigeonhole principle, at least 4 of these 3-tuples must belong to the same group among the 13 groups. Writing these 4 3-tuples as a $3 \times 4$ number table, the 3 rows of numbers are $S, T, U$ which satisfy the requirements of the problem.
In summary, the smallest natural number $n=982$.