29.2.3 For any two elements $a=\left(a_{1}, a_{2}, \cdots, a_{2^{n}}\right)$ and $b=\left(b_{1}, b_{2}, \cdots, b_{2}{ }^{n}\right)$ in $S_{n}$, let
$$
d(a, b)=\sum_{i=1}^{i^{*}}\left|a_{i}-b_{i}\right| .
$$

If $A \subseteq S_{n}$, and for any two different elements $a, b$ in $A$, we have $d(a, b) \geqslant 2^{n-1}$, then $A$ is called a good subset. Find the maximum number of elements in a good subset of $S_{u}$.
If $a_{i} \neq b_{i}$, then $a$ and $b$ are said to have a different component at the $i$-th position. It is easy to see that $d(a, b)$ is the number of different component pairs in $a$ and $b$. If $A \subseteq S_{n}$, and any two different elements in $A$ contain at least $2^{n-1}$ different component pairs, then $A$ is called $n$-level good.
First, prove: If $A$ is $n$-level good, then $|A| \leqslant 2^{n+1}$.
In fact, if $|A| \geqslant 2^{n+1}+1$, then $A$ has at least $2^{n}+1$ elements with the same last bit. Without loss of generality, let $B \subseteq A, |B|=2^{n}+1$, and each element in $B$ has the last bit as 0.
Consider the total number of different component pairs $N$ among all elements in $B$.
On one hand, since any two different elements in $B$ have at least $2^{n-1}$ different component pairs, we have
$$
N \geqslant 2^{n-1} \cdot \mathrm{C}_{2^{n}+1}^{2} - 2^{n-1} \cdot 2^{n-1} \cdot (2^{n}+1).
$$

On the other hand, for the $i$-th component of elements in $B$ ($1 \leqslant i \leqslant 2^{n}-1$), let there be $x_{i}$ zeros and $2^{n}+1-x_{i}$ ones, then
$$
\begin{aligned}
N & = \sum_{i=1}^{2^{n}-1} x_{i} (2^{n}+1 - x_{i}) \leqslant \sum_{i=1}^{2^{n}-1} 2^{n-1} \cdot (2^{n-1} + 1) \\
& - 2^{n-1} (2^{n-1} + 1) (2^{n} - 1),
\end{aligned}
$$

Thus, $2^{n-1} \cdot (2^{n-1} + 1) \cdot (2^{n} - 1) \geqslant 2^{n-1} \cdot 2^{n-1} \cdot (2^{n} + 1)$, hence $0 \leqslant -1$, which is a contradiction!
Next, recursively construct an $n$-level good subset $A_{n}$ such that $|A_{n}| = 2^{n+1}$.
When $n=1$, take $A_{1} = \{(0,0), (0,1), (1,0), (1,1)\}$.
Assume $A_{n}$ is an $n$-level good subset, and $|A_{n}| = 2^{n+1}$. For $a \in S_{n}, a = (a_{1}, a_{2}, \cdots, a_{2}^{n})$, define
$$
a' = (1-a_{1}, 1-a_{2}, \cdots, 1-a_{2^{n}})_{\text{+ let }} A_{n+1} = \{a, a\} \text{ or } \{a, a'\} \mid a \in A_{n} \subseteq S_{n+1},
$$

Clearly,
$$
|A_{n+1}| = 2 |A_{n}| = 2^{n+2}.
$$

For different elements $a, b$ in $A_{n}$, it is easy to see:
$$
\begin{aligned}
d((b, b), (a, a')) & = 2^{n}, d((a, a), (b, b)) \\
& = 2 d(a, b) \geqslant 2 \cdot 2^{n-1} = 2^{n}, \\
d((a, a), (b, b')) & = 2^{n}, d((a, a'), (b, b')) \\
& = 2 d(a, b) \geqslant 2 \cdot 2^{n-1} = 2^{n},
\end{aligned}
$$

Therefore, $A_{n+1}$ is $(n+1)$-level good.
Thus, the maximum value of $|A|$ is $2^{n+1}$.