The 90th problem: Let $\mathrm{n}$ be a positive integer, $\mathrm{S}_{\mathrm{n}}=\left\{\left(\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{2^{n}}\right) \mid \mathrm{a}_{\mathrm{i}} \in\{0,1\}, 1 \leq \mathrm{i} \leq 2^{\mathrm{n}}\right\}$, for $\mathrm{a}, \mathrm{~b} \in$ $S_{n}, a=\left(a_{1}, a_{2}, \ldots, a_{2^{n}}\right), \quad b=\left(b_{1}, b_{2}, \ldots, b_{2^{n}}\right)$, define $d(a, b)=\sum_{i=1}^{2^{n}}\left|a_{i}-b_{i}\right|$. If for any a, $b \in A, d(a, b) \geq 2^{n-1}$, then $A \subseteq S_{n}$ is called a "good subset". Find the maximum possible value of $|A|$.
The 90th problem,
Solution: Let set $\mathrm{A}$ have $\mathrm{k}$ elements whose first component is 1, denoted as $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{\mathrm{k}}$. Let
$$
\mathrm{T}=\sum_{1 \leq \mathrm{i}<\mathrm{j} \leq \mathrm{k}} \mathrm{d}\left(\alpha_{\mathrm{i}}, \quad \alpha_{\mathrm{j}}\right)
$$

Then, according to the condition, $\mathrm{T} \geq \mathrm{C}_{\mathrm{k}}^{2} \cdot 2^{\mathrm{n}-1}$. For the $\mathrm{i}$-th component of elements $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{k}$, let there be $x_{i}$ ones and $k-x_{i}$ zeros $\left(i=1, 2, \ldots, 2^{n}\right)$, where $\mathrm{x}_{1}=\mathrm{k}$. Then
$$
T=\sum_{i=1}^{2^{n}} x_{i}\left(k-x_{i}\right)=\sum_{i=2}^{2^{n}} x_{i}\left(k-x_{i}\right) \leq \frac{k^{2}}{4} \cdot\left(2^{n}-1\right) \Rightarrow k \leq 2^{n}
$$

Similarly, the number of elements in set $A$ whose first component is 0 is also at most $2^{n}$. Therefore, $|A| \leq 2^{n}+2^{n}=$ $2^{\mathrm{n}+1}$.
Below, we use induction to construct an example.
When $\mathrm{n}=1$, $\mathrm{A}_{1}=\{(0,0),(0,1),(1,0),(1,1)\}$ satisfies the condition.
Assume that when $\mathrm{n}$, $\mathrm{A}_{\mathrm{n}}=\left\{\alpha_{1}, \alpha_{2}, \ldots, \alpha_{2^{\mathrm{n}+1}}\right\}$ satisfies the condition, where $\alpha_{\mathrm{i}}=\left(\mathrm{a}_{1}^{(i)}, \mathrm{a}_{2}^{(i)}, \ldots, \mathrm{a}_{2^{\mathrm{n}}}^{(i)}\right)(\mathrm{i}=$ $\left.1,2, \ldots, 2^{n+1}\right)$. At this point, $d\left(\alpha_{i}, \alpha_{j}\right) \geq 2^{n-1}\left(1 \leq i, j \leq 2^{n+1}, i \neq j\right)$.
When $n+1$, take $A_{n+1}=\left\{\beta_{1}, \beta_{2}, \ldots, \beta_{2^{n+1}}, \gamma_{1}, \gamma_{2}, \ldots, \gamma_{2^{n+1}}\right\}$, where
$$
\begin{array}{l}
\beta_{\mathrm{i}}=\left(a_{1}^{(i)}, a_{2}^{(i)}, \ldots, a_{2^{n}}^{(i)}, a_{1}^{(i)}, a_{2}^{(i)}, \ldots, a_{2^{n}}^{(i)}\right) \\
c_{\mathrm{i}}=\left(a_{1}^{(i)}, a_{2}^{(i)}, \ldots, a_{2^{n}}^{(i)}, 1-a_{1}^{(i)}, 1-a_{2}^{(i)}, \ldots, 1-a_{2^{n}}^{(i)}\right)
\end{array}
$$

It can be proven that the constructed $\mathrm{A}_{\mathrm{n}+1}$ satisfies the condition.
In fact, for any $\mathrm{i} \neq \mathrm{j}$, we have
$$
\begin{array}{l}
\mathrm{d}\left(\beta_{\mathrm{i}}, \beta_{\mathrm{j}}\right)=2 \mathrm{~d}\left(\alpha_{\mathrm{i}}, \alpha_{\mathrm{i}}\right) \geq 2^{\mathrm{n}} \\
\mathrm{d}\left(\gamma_{\mathrm{i}}, \gamma_{\mathrm{j}}\right)=2 \mathrm{~d}\left(\alpha_{\mathrm{i}}, \alpha_{\mathrm{j}}\right) \geq 2^{\mathrm{n}} \\
\mathrm{d}\left(\beta_{\mathrm{i}}, \gamma_{\mathrm{j}}\right)=2^{\mathrm{n}}+\mathrm{d}\left(\alpha_{\mathrm{i}}, \alpha_{\mathrm{j}}\right)-\mathrm{d}\left(\alpha_{\mathrm{i}}, \alpha_{\mathrm{j}}\right)=2^{\mathrm{n}} \\
\mathrm{d}\left(\beta_{\mathrm{i}}, \gamma_{\mathrm{i}}\right)=2^{\mathrm{n}} \\
\end{array}
$$

Therefore, the constructed $A_{n+1}$ satisfies the condition, and $\left|A_{n+1}\right|=2^{n+2}$.
In summary, the maximum value of $|A|$ that satisfies the condition is $2^{n+1}$.