$11 \cdot 29$ Find the smallest positive integer $n$, such that for any given $n$ rectangles with side lengths being positive integers not exceeding 100, there always exist 3 rectangles $R_{1}, R_{2}, R_{3}$, such that $R_{1}$ can be placed inside $R_{2}$, and $R_{2}$ can be placed inside $R_{3}$.
[Solution] Map each rectangle to an integer point $(x, y)$ on the coordinate plane, where $x$ and $y$ are the width and length of the rectangle, respectively, with $1 \leqslant x \leqslant y \leqslant 100$.
Consider the following two sets:
$$
\begin{array}{l}
S=\{(i, 100-i+1) \mid i=1,2, \cdots, 50\}, \\
T=\{(i, 100-i) \mid i=1,2, \cdots, 50\} .
\end{array}
$$

Each of these sets contains 50 elements, and the 50 integer points in each set correspond to rectangles that do not contain each other (meaning one cannot be placed inside another). Therefore, if we select 3 rectangles from the 100 rectangles corresponding to the integer points in these two sets, there must be two rectangles in the same set, and these two rectangles will not meet the requirements of the problem. This indicates that the smallest positive integer $n \geqslant 101$.

When there are 101 rectangles, consider the following 50 sets of integer points, each lying on a zigzag line:
$$
\begin{array}{l} 
M_{i}=\{(i, i),(i+1, i), \cdots,(100-i+1, i),(100-i+1, i+1), \\
\cdots,(100-i+1,100-i+1)\}, i=1,2, \cdots, 50 .
\end{array}
$$

It is easy to see that $\left|M_{i}\right|=203-4 i$ and any two integer points in each $M_{i}$ correspond to rectangles where the smaller can be placed inside the larger. Given 101 rectangles, they are distributed among these 50 sets. By the pigeonhole principle, there must be 3 rectangles whose corresponding integer points belong to the same set, and these 3 rectangles will meet the requirements of the problem.
In summary, the smallest positive integer $n=101$.