The twelfth question: A company has $n(n \geq 2)$ employees, and now everyone is required to vote for an outstanding employee, with each person casting exactly one vote to another employee. Try to find the maximum value of $\mathrm{k}$, such that no matter how the voting goes, it is always possible to select $\mathrm{k}$ people, none of whom have voted for any of the others among these $\mathrm{k}$ people.

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The 12th question:
Solution: Use $n$ points to represent these $n$ employees. If employee $a$ has voted for employee $b$, then draw a directed edge $\overrightarrow{b}$, resulting in an $n$-order directed graph $D$. Clearly, the out-degree of each point in $D$ is 1, so $|E(D)|=n$.

Now consider the underlying graph $G$ of $D$, then $|E(G)|=|E(D)|=n$. According to Turán's theorem, the independence number of graph $G$, $\beta(G) \geq \sum_{v \in V(G)} \frac{1}{1+d_{G}(V)} \geq \frac{n^{2}}{n+\sum_{v \in V(G)} d_{G}(V)}=\frac{n^{2}}{n+2n}=\frac{n}{3}$, hence there exists a $\left\lfloor \frac{n}{3} \right\rfloor$-order independent set in graph $G$. These $\left\lfloor \frac{n}{3} \right\rfloor$ points correspond to employees who have not voted for each other.

On the other hand, let the $n$ employees be $A_{1}, A_{2}, \ldots, A_{n}$. Let $n=3m+r, m=\left\lceil\frac{n}{3}\right\rceil-1$, then $1 \leq r \leq 3$. We divide these $n$ employees into $m+1=\left\lfloor \frac{n}{3} \right\rfloor$ groups, the first $m$ groups are $\left\{A_{3i-2}, A_{3i-1}, A_{3i}\right\} (1 \leq i \leq m)$, and the $(m+1)$-th group is $\left\{A_{3m+1}, A_{3m+2}, \ldots, A_{3m+r}\right\}$.
The voting situation in the $i$-th group $(1 \leq i \leq m)$ is as follows: $A_{3i-2}$ votes for $A_{3i-1}$, $A_{3i-1}$ votes for $A_{3i}$, and $A_{3i}$ votes for $A_{3i-2}$.
Now provide the voting scheme for the $(m+1)$-th group: When $r=1$, $A_{3m+1}$ votes for $A_{3m}$; when $r=2$, $A_{3m+1}$ and $A_{3m+2}$ vote for each other; when $r=3$, $A_{3m+1}$ votes for $A_{3m+2}$, $A_{3m+2}$ votes for $A_{3m+3}$, and $A_{3m+3}$ votes for $A_{3m+1}$.

In the above voting scheme, according to the requirements of the problem, at most one person can be selected from each group, so at most $\left\lceil \frac{n}{3} \right\rceil$ can be selected.
In summary, the maximum value of $k$ is $\left\lfloor \frac{n}{3} \right\rfloor$.