11. (20 points) Given positive real numbers $0<a<b$, let $x_{1}$, $x_{2}$, $x_{3}$, $x_{4} \in [a, b]$. Find
$$
\frac{\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\frac{x_{3}^{2}}{x_{4}}+\frac{x_{4}^{2}}{x_{1}}}{x_{1}+x_{2}+x_{3}+x_{4}}
$$

the minimum and maximum values.
11. (1) Given $x_{1}, x_{2}, x_{3}, x_{4} \in [a, b], 0 < a < b$,

we know
$$
\begin{array}{l}
\frac{x_{1}^{2}}{x_{2}} + x_{2} \geqslant 2 \sqrt{\frac{x_{1}^{2}}{x_{2}} \cdot x_{2}} = 2 x_{1}, \\
\frac{x_{2}^{2}}{x_{3}} + x_{3} \geqslant 2 \sqrt{\frac{x_{2}^{2}}{x_{3}} \cdot x_{3}} = 2 x_{2}, \\
\frac{x_{3}^{2}}{x_{4}} + x_{4} \geqslant 2 \sqrt{\frac{x_{3}^{2}}{x_{4}} \cdot x_{4}} = 2 x_{3}, \\
\frac{x_{4}^{2}}{x_{1}} + x_{1} \geqslant 2 \sqrt{\frac{x_{4}^{2}}{x_{1}} \cdot x_{1}} = 2 x_{4}. \\
\text{Thus, } \frac{x_{1}^{2}}{x_{2}} + \frac{x_{2}^{2}}{x_{3}} + \frac{x_{3}^{2}}{x_{4}} + \frac{x_{4}^{2}}{x_{1}} \geqslant x_{1} + x_{2} + x_{3} + x_{4}, \text{ and }
\end{array}
$$

the equality holds if and only if $x_{1} = x_{2} = x_{3} = x_{4}$. 
Therefore, when $x_{1} = x_{2} = x_{3} = x_{4}$, the minimum value is 1.
(2) Also, from the problem statement, we know
$$
\frac{x_{i}}{x_{i+1}} \in \left[\frac{a}{b}, \frac{b}{a}\right] \left(x_{5} = x_{1}\right).
$$

Notice that, $\frac{\sqrt{\frac{x_{i}^{2}}{x_{i+1}}}}{\sqrt{x_{i+1}}} = \frac{x_{i}}{x_{i+1}}$.
Thus, $\frac{a}{b} \leqslant \frac{\sqrt{\frac{x_{i}^{2}}{x_{i+1}}}}{\sqrt{x_{i+1}}} \leqslant \frac{b}{a}$.
$\Rightarrow \left(\sqrt{\frac{x_{i}^{2}}{x_{i+1}}} - \frac{a}{b} \sqrt{x_{i+1}}\right)$.
$\left(\sqrt{\frac{x_{i}^{2}}{x_{i+1}}} - \frac{b}{a} \sqrt{x_{i+1}}\right) \leqslant 0$.
$$
\Rightarrow \frac{x_{i}^{2}}{x_{i+1}} - \left(\frac{a}{b} + \frac{b}{a}\right) x_{i} + x_{i+1} \leqslant 0.
$$

Therefore, for all $1 \leqslant i \leqslant 4$,
$$
\begin{array}{l}
\frac{x_{i}^{2}}{x_{i+1}} \leqslant \left(\frac{a}{b} + \frac{b}{a}\right) x_{i} - x_{i-1} \left(x_{5} = x_{1}\right). \\
\text{Hence, } \frac{x_{1}^{2}}{x_{2}} + \frac{x_{2}^{2}}{x_{3}} + \frac{x_{3}^{2}}{x_{4}} + \frac{x_{4}^{2}}{x_{1}} \\
\leqslant \left(\frac{b}{a} + \frac{a}{b} - 1\right) \left(x_{1} + x_{2} + x_{3} + x_{4}\right) \\
\Rightarrow \frac{\frac{x_{1}^{2}}{x_{2}} + \frac{x_{2}^{2}}{x_{3}} + \frac{x_{3}^{2}}{x_{4}} + \frac{x_{4}^{2}}{x_{1}}}{x_{1} + x_{2} + x_{3} + x_{4}} \leqslant \frac{b}{a} + \frac{a}{b} - 1, \\
\end{array}
$$

the equality holds if and only if $x_{1} = a, x_{2} = b, x_{3} = a, x_{4} = b$ or $x_{1} = b, x_{2} = a, x_{3} = b, x_{4} = a$.
Therefore, the maximum value is $\frac{b}{a} + \frac{a}{b} - 1$.
In summary, the minimum and maximum values are 1 and $\frac{b}{a} + \frac{a}{b} - 1$, respectively.