18. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, satisfying $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}=n$, find the minimum value of $x_{1}+\frac{x_{2}^{2}}{2}+$ $\frac{x_{3}^{2}}{3}+\cdots+\frac{x_{n}^{n}}{n}$. (1995 Polish Mathematical Olympiad Problem)
18. From the known $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}=n \geqslant n \sqrt[n]{\frac{1}{x_{1}} \cdot \frac{1}{x_{2}} \cdots \cdots \cdot \frac{1}{x_{n}}}$, we get $x_{1} x_{2} x_{3} \cdots x_{n} \geqslant 1$. To $\frac{x_{1}+\frac{x_{2}^{2}}{2}+\frac{x_{3}^{2}}{3}+\cdots+\frac{x_{n}^{n}}{n}}{1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}} \geqslant\left(x_{1} x_{2} x_{3} \cdots x_{n}\right)^{p}=1$. Where $p=\frac{1}{1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}}$. Therefore, when $x_{1}=x_{2}=x_{3}=\cdots=x_{n}=1$, $x_{1}+\frac{x_{2}^{2}}{2}+\frac{x_{3}^{2}}{3}+\cdots+\frac{x_{n}^{n}}{n}$ takes the minimum value $1+\frac{1}{2}+$ $\frac{1}{3}+\cdots+\frac{1}{n}$