Three. (25 points) Given $n(n>1)$ integers (which can be the same) $x_{1}, x_{2}, \cdots, x_{n}$ satisfy
$$
x_{1}+x_{2}+\cdots+x_{n}=x_{1} x_{2} \cdots x_{n}=9111 \text {. }
$$

Find the maximum value of $x_{1}, x_{2}, \cdots, x_{n}$ when $n$ takes the minimum value.
Three, from $x_{1} x_{2} \cdots x_{n}=9111$, we know that $x_{1}, x_{2}, \cdots, x_{n}$ are all odd numbers.

Since $x_{1}+x_{2}+\cdots+x_{n}=9111$ is odd, $n (n>1)$ must be odd.
If $n=3$, then
$$
x_{1}+x_{2}+x_{3}=x_{1} x_{2} x_{3}=9111 \text {. }
$$

By the symmetry of the condition equations, without loss of generality, assume
$$
x_{1} \geqslant x_{2} \geqslant x_{3} \text {. }
$$

Then $x_{1}+x_{2}+x_{3} \leqslant x_{1}+x_{1}+x_{1}$. Thus,
$$
x_{1} \geqslant \frac{x_{1}+x_{2}+x_{3}}{3}=3037, x_{2} x_{3} \leqslant 3 \text {. }
$$

If $x_{1}=3037$, then $x_{2} x_{3}=3$.
Thus, $x_{2}+x_{3} \leqslant 4$. Therefore,
$$
x_{1}+x_{2}+x_{3} \leqslant 3037+4=3041 < 9111$.
If $x_{1} > 3037$, then $x_{1} \geqslant 9111$, which implies $x_{2} x_{3}=1$ and $x_{2}+x_{3}=0$, which is impossible.
Therefore, $n \geqslant 5$. Also, because
$$
\begin{array}{l}
9111+1+1+(-1)+(-1) \\
=9111 \times 1 \times 1 \times(-1) \times(-1)=9111,
\end{array}
$$
the minimum value of $n$ is 5.
When $n=5$,
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}+x_{5} \\
=x_{1} x_{2} x_{3} x_{4} x_{5}=9111,
\end{array}
$$
the maximum value among $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ is 9111.
Below, we prove by contradiction that the maximum value among $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ is 9111.

Assume there is a number greater than 9111 among $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, without loss of generality, let $x_{1}>9111$. Then
$$
\left|x_{1}\right|>9111 \text {. }
$$

Clearly, $\left|x_{2} x_{3} x_{4} x_{5}\right|>0$. Thus,
$$
\left|x_{1}\right| \cdot\left|x_{2} x_{3} x_{4} x_{5}\right|>9111\left|x_{2} x_{3} x_{4} x_{5}\right| \text {. }
$$

Therefore, $9111>9111\left|x_{2} x_{3} x_{4} x_{5}\right|$.
Thus, $\left|x_{2} x_{3} x_{4} x_{5}\right|<1$.
This contradicts the fact that $x_{2}, x_{3}, x_{4}, x_{5}$ are all odd numbers.
Hence, when $n$ takes the minimum value 5, the maximum value among $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ is 9111.
(Zhou Dongting, Experimental Middle School of Xishui County, Huanggang City, Hubei Province, 438200 Fang Chengsheng, No.1 Middle School of Xishui County, Huanggang City, Hubei Province, 438200)