13. A. For any real number $x$, let $\lfloor x\rfloor$ denote the greatest integer not exceeding the real number $x$, $\{x\}=x-\lfloor x\rfloor$, and $\lceil x\rceil$ denote the smallest integer not less than the real number $x$. If $x_{1}, x_{2}, \cdots, x_{m}$ $\left(0 \leqslant x_{1}<x_{2}<\cdots<x_{m} \leqslant 10\right)$ are all the real numbers between 0 and 10 that satisfy the equation $\lfloor x\rfloor\{x\}\lceil x\rceil=1$, find the value of $x_{1}+$ $x_{2}+\cdots+x_{m}$.
13. A. Let the real number $x(0 \leqslant x \leqslant 10)$ satisfy the equation $\lfloor x\rfloor\{x\}\lceil x\rceil=1$.
Obviously, $\{x\} \neq 0$.
Therefore, $x$ is not an integer.
At this point, if we denote $\lfloor x\rfloor=a(0 \leqslant a \leqslant 9)$, then $a<x<a+1$.
Thus, $\lceil x\rceil=a+1$.
Also, $a, a+1 \neq 0$, and
$\{x\}=\frac{1}{\lfloor x\rfloor\lceil x\rceil}=\frac{1}{a(a+1)}$,
so $x=a+\{x\}=a+\frac{1}{a(a+1)}$.
When $a$ takes the values $1,2, \cdots, 9$, the obtained $x$ values, when arranged from smallest to largest, are all the solutions, thus, $m=9$, and
$$
\begin{array}{l}
x_{k}=k+\frac{1}{k(k+1)}(k=1,2, \cdots, 9) . \\
\text { Hence } \sum_{k=1}^{9} x_{k}=\sum_{k=1}^{9}\left[k+\frac{1}{k(k+1)}\right] \\
=45+1-\frac{1}{10}=\frac{459}{10} .
\end{array}
$$