Question 93,
Answer: (1) First, prove: when $n=40$, there exists a scenario such that no matter how the groups are formed, each class will have at least 10 tall students.
Indeed, arrange all $3 \times 40=120$ students by height from tallest to shortest.
Let the 1st to 10th and 91st to 120th be in Class 1; the 11th to 30th and 71st to 90th be in Class 2; the 31st to 70th be in Class 3. Then, no matter how the groups are formed, the 1st to 10th are all tall; among the 11th to 30th, at least $20-10=10$ are tall; among the 31st to 70th, at least $40-(10+20)=10$ are tall. Thus, each class will have at least 10 tall students.
(2) Next, prove: when $n=39$, no matter what the scenario, there always exists a grouping scheme such that at least one class has fewer than 10 tall students.

Indeed, let the heights of 39 people in Class 1 from tallest to shortest be: $\mathrm{a}_{1}>\mathrm{a}_{2}>\ldots>\mathrm{a}_{39}$; the heights of 39 people in Class 2 from tallest to shortest be: $b_{1}>b_{2}>\ldots>b_{39}$; the heights of 39 people in Class 3 from tallest to shortest be: $c_{1}>c_{2}>\ldots>$ $c_{39}$. Assume $a_{20}=\max \left\{a_{20}, b_{20}, c_{20}\right\}$. Then group as follows:
$$
\begin{array}{l}
\left(a_{1}, b_{20}, c_{20}\right),\left(a_{2}, b_{21}, c_{21}\right), \cdots,\left(a_{20}, b_{39}, c_{39}\right) ; \\
\left(a_{21}, b_{1}, c_{1}\right),\left(a_{22}, b_{2}, c_{22}\right), \cdots,\left(a_{39}, b_{19}, c_{19}\right) .
\end{array}
$$

Then, in Class 1, $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{20}$ are all tall, so in Classes 2 and 3, there are at most 19 tall students in total. According to the pigeonhole principle, there must be one class with at most 9 tall students.
In summary, the smallest value of $\mathrm{n}$ that satisfies the condition is 40.
Note: The problem can be generalized as follows: Suppose there are $\mathrm{k}$ classes, each with $\mathrm{n}$ students, and the heights of these $\mathrm{kn}$ students are all different. Now, these students are divided into $\mathrm{n}$ groups, each group consisting of $\mathrm{k}$ students from different classes, and the tallest student in each group is called a "tall student." Given that no matter how the groups are formed, each class will have at least $\mathrm{m}$ "tall students," find the smallest possible value of $\mathrm{n}$.
It can be proven: the smallest value of $\mathrm{n}$ is $\mathrm{m} \cdot 2^{\mathrm{k}-1}$.
On one hand, arrange all students by height from tallest to shortest. Let the first $\mathrm{m}$ be in Class 1; the next $2 \mathrm{~m}$ be in Class 2; the next $2^{2} \mathrm{~m}$ be in Class 3; the next $2^{3} \mathrm{~m}$ be in Class 4; $\cdots$; the next $2^{\mathrm{k}-1} \mathrm{~m}$ be in Class $\mathrm{k}$. Then, arbitrarily assign the remaining students to classes, ensuring each class has exactly $2^{\mathrm{k}-1} \mathrm{~m}$ students. In this scenario, no matter how the groups are formed, each class will have at least $\mathrm{m}$ tall students.

On the other hand, we prove by induction on $\mathrm{k}$: when $n=2^{k-1} m-1$, no matter what the scenario, there always exists a grouping scheme such that at least one class has fewer than $\mathrm{m}$ tall students.
When $\mathrm{k}=1, 2$, the conclusion is obviously true.
Assume the conclusion holds for $\mathrm{k}$. Consider the case for $\mathrm{k}+1$. At this point, each class has exactly $2^{\mathrm{k}} \mathrm{m}-1$ students. For any $\mathrm{i} \in\{1,2, \ldots, \mathrm{k}+1\}$, let the heights of the $2^{\mathrm{k}} \mathrm{m}-1$ students in the $\mathrm{i}$-th class from tallest to shortest be:
$$
\begin{array}{l}
\mathrm{a}_{1}^{(\mathrm{i})}>\mathrm{a}_{2}^{(\mathrm{i})}>\ldots>\mathrm{a}_{2^{(\mathrm{k}} \mathrm{m}-1}^{(1)} \\
\text { Assume } a_{2^{k-1} m}^{(k+1)}=\max \left\{a_{2^{k-1} m}^{(1)}, a_{2^{k-1} m}^{(2)}, \ldots, a_{2^{k-1} m}^{(k+1)}\right\} \text { . Then group as follows: } \\
\left(a_{2^{k-1} m}^{(1)}, \quad a_{2^{k-1} m}^{(2)}, \ldots, a_{2^{k-1} m}^{(k)}, a_{1}^{(k+1)}\right) \\
\left(a_{2^{k-1} m+1}^{(1)}, a_{2^{k-1} m+1}^{(2)}, \ldots, a_{2^{k-1} m+1}^{(k)}, a_{2}^{(k+1)}\right) \\
\left(a_{2^{k-1} m+2}^{(1)}, \quad a_{2^{k-1} m+2}^{(2)}, \ldots, a_{2^{k-1} m+2}^{(k)}, a_{3}^{(k+1)}\right) \\
\left(a_{2^{k} m-1}^{(1)}, a_{2^{k} m-1}^{(2)}, \ldots, a_{2^{k} m-1}^{(k)}, a_{2^{k-1} m}^{(k+1)}\right) \\
\end{array}
$$

Then, in the $(k+1)$-th class, $a_{1}^{(k+1)}, a_{2}^{(k+1)}, \ldots, a_{2^{k-1} m}^{(k+1)}$ are all tall.
Now consider the first $1 \sim \mathrm{k}$ classes, each with the first $2^{\mathrm{k}-1} \mathrm{~m}-1$ and the last $2^{\mathrm{k}-1} \mathrm{~m}-1$ students in the $(k+1)$-th class. According to the induction hypothesis, there exists a grouping scheme such that, without the students in the $(k+1)$-th class, at least one of the first $\mathrm{k}$ classes has at most $\mathrm{m}-1$ tall students. Now, add the students from the $(k+1)$-th class, and the number of tall students in that class will not increase, still at most $m-1$ tall students.
In summary, when there are $\mathrm{k}$ classes, the smallest value of $\mathrm{n}$ is $\mathrm{m} \cdot 2^{\mathrm{k}-1}$.
Question 93: There are three classes, each with exactly $\mathrm{n}$ students, and the heights of these $3 \mathrm{n}$ students are all different. Now, these students are divided into $\mathrm{n}$ groups, with 3 students in each group coming from different classes, and the tallest student in each group is called a “tall person.” It is known that no matter how the groups are formed, each class has at least 10 “tall persons.” Find the minimum possible value of $\mathrm{n}$.