Question 81: Find all positive integers $\mathrm{n}$ and $\mathrm{n}$-degree polynomials $\mathrm{f}(\mathrm{x})=\sum_{\mathrm{i}=0}^{\mathrm{n}} \mathrm{a}_{\mathrm{i}} \mathrm{x}^{\mathrm{i}}$, such that $\left\{\mathrm{a}_{0}, \mathrm{a}_{1}, \ldots, \mathrm{a}_{\mathrm{n}}\right\}=$ $\{0,1, \ldots, \mathrm{n}\}$, and all $\mathrm{n}$ roots of $\mathrm{f}(\mathrm{x})$ are real.
Question 81: Find all positive integers $\mathrm{n}$ and $\mathrm{n}$-degree polynomials $\mathrm{f}(\mathrm{x})=\sum_{\mathrm{i}=0}^{\mathrm{n}} \mathrm{a}_{\mathrm{i}} \mathrm{x}^{\mathrm{i}}$, such that $\left\{\mathrm{a}_{0}, \mathrm{a}_{1}, \ldots\right.$, $\left.a_{n}\right\}=\{0,1, \ldots, n\}$, and all $n$ roots of $f(x)$ are real numbers.
Solution: When $\mathrm{n}=1$, it is clear that only $\mathrm{f}(\mathrm{x})=\mathrm{x}$ satisfies the condition.
Assume $n \geq 2$, and let the $n$ roots of $f(x)$ be $x_{1}, x_{2}, \ldots, x_{n}$.
(1) First, prove: $a_{0}=0$.
If $a_{0} \neq 0$, then $x_{1}, x_{2}, \ldots, x_{n}$ are all negative real numbers. According to Vieta's formulas, we have: $\sum_{i=1}^{n}\left|x_{i}\right|=\frac{a_{n-1}}{a_{n}}$, $\prod_{\mathrm{i}=1}^{\mathrm{n}}\left|\mathrm{x}_{\mathrm{i}}\right|=\frac{\mathrm{a}_{0}}{\mathrm{a}_{\mathrm{n}}}$. By the AM-GM inequality, we have:
$$
\begin{array}{l}
\left(\frac{\sum_{i=1}^{n} \mid x_{i} \mid}{n}\right)^{n} \geq \prod_{i=1}^{n}\left|x_{i}\right| \\
\Rightarrow\left(\frac{a_{n-1}}{n a_{n}}\right)^{n} \geq \frac{a_{0}}{a_{n}} \\
\Rightarrow a_{n-1}^{n} \geq n^{n} \cdot a_{n}^{n-1} \cdot a_{0}>n^{n}
\end{array}
$$

This contradicts $a_{n-1} \in\{0,1, \ldots, n\}$. Therefore, $a_{0}=0$, and thus $x_{1}, x_{2}, \ldots, x_{n}$ include exactly one 0 and the rest are negative real numbers.

Assume $x_{n}=0$, then $\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}=\{1,2, \ldots, n\}$, and $x_{1}, x_{2}, \ldots, x_{n-1}$ are the roots of $g(x)=\sum_{i=1}^{n} a_{i} x^{i-1}$.
(2) Next, prove: $\mathrm{n} \leq 3$, and $\left\{\mathrm{a}_{1}, \mathrm{a}_{\mathrm{n}}\right\}=\{1,2\}$.
In fact, at this point, $g(x)=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{i}} \mathrm{x}^{\mathrm{i}-1}=a_{n} \prod_{i=1}^{n-1}\left(x+\left|x_{i}\right|\right)$, so
$$
\begin{array}{l}
g(1)=\sum_{i=1}^{n} a_{i}=\frac{n(n+1)}{2}=a_{n} \prod_{i=1}^{n-1}\left(1+\left|x_{i}\right|\right) \geq a_{n} \cdot 2^{n-1} \cdot \sqrt{\prod_{i=1}^{n-1}\left|x_{i}\right|}=a_{n} \cdot 2^{n-1} \cdot \sqrt{\frac{a_{1}}{a_{n}}} \\
\Rightarrow \frac{n(n+1)}{2} \geq 2^{n-1} \cdot \sqrt{a_{1} a_{n}} \geq 2^{n-1} \cdot \sqrt{2}
\end{array}
$$

Thus, $n \leq 3$, and it must be that $a_{1} a_{n}=2$, i.e., $\left\{a_{1}, a_{n}\right\}=\{1,2\}$.
Upon verification, when $n=2$, $f(x)=x^{2}+2 x$ and $f(x)=2 x^{2}+x$ both satisfy the condition; when $n=3$, $f(x)=x^{3}+3 x^{2}+2 x$ and $f(x)=2 x^{3}+3 x^{2}+x$ both satisfy the condition.

In summary, the polynomials that satisfy the conditions are $f(x)=x, f(x)=x^{2}+2 x, f(x)=2 x^{2}+x$, $f(x)=x^{3}+3 x^{2}+2 x$, and $f(x)=2 x^{3}+3 x^{2}+x$.