Question 230, Let $S$ be the set of ordered 7-tuples $\left(a_{1}, a_{2}, \ldots, a_{7}\right)$, where $a_{i}=1$ or 0. For any two elements $a=\left(a_{1}, a_{2}, \ldots, a_{7}\right)$ and $b=\left(b_{1}, b_{2}, \ldots, b_{7}\right)$ in $S$, define the distance between $a$ and $b$ as $\sum_{i=1}^{7}\left|a_{i}-b_{i}\right|$. The set $T$ is a subset of $S$, where the distance between any two elements is at least 3. Find the maximum value of $|T|$.
Question 230: Proof: According to the given conditions, it is required that any two elements in the set $\mathrm{T}$ differ in at least three positions.
We first prove that $|\mathrm{T}| \leq 16$.
Otherwise, if $|T| \geq 17$. Note that the first three positions of elements in $T$ have $2^{3}=8$ different combinations. According to the pigeonhole principle, there are at least $\left\lceil\frac{17}{8}\right]=3$ elements in the set $\mathrm{T}$ whose first three positions are identical. Let these three elements be $a$, $b$, and $c$. Consider the last four positions of the digits of $a$, $b$, and $c$.

If the last four positions of $a$ and $b$ are all different, then $c$ and $a$ differ in at least three positions, which means $c$ and $b$ differ in at most one position, leading to a contradiction; if the last four positions of $a$ and $b$ differ in exactly three positions, then $c$ and $a$ differ in at least three positions, which means $c$ and $b$ differ in at most two positions, leading to a contradiction.
Thus, we know that $|T| \leq 16$. Below, we provide an example where $|T|=16$.
$(0,0,0,0,0,0,0),(1,1,1,0,0,0,0),(1,0,0,1,1,0,0)$,
$(1,0,0,0,0,1,1),(0,1,0,1,0,1,0),(0,1,0,0,1,0,1)$,
$(0,0,1,1,0,0,1),(0,0,1,0,1,1,0),(0,0,0,1,1,1,1)$,
$(0,1,1,0,0,1,1),(0,1,1,1,1,0,0),(1,0,1,0,1,0,1)$,
$(1,0,1,1,0,1,0),(1,1,0,0,1,1,0),(1,1,0,1,0,0,1)$,
$(1,1,1,1,1,1,1)$.
In conclusion, the maximum value of $|\mathrm{T}|$ is 16.