【Solution】Solution: According to the analysis, the sum of the digits of $\overline{\mathrm{abc}}$ and $\overline{\mathrm{cba}}$ are both: $a+b+c$, the remainder when the three-digit numbers $\overline{\mathrm{abc}}$ and $\overline{\mathrm{cba}}$ are divided by the sum of their digits is 1,

According to the principle of remainders, the difference between $\overline{\mathrm{abc}}$ and $\overline{\mathrm{cba}}$ can be divided by $a+b+c$. Let $\overline{\mathrm{abc}}-\overline{\mathrm{cba}}=\mathrm{n}(\mathrm{a}+\mathrm{b}+\mathrm{c})$. According to the principle of place value, we have:
$$
\begin{array}{l}
\overline{\mathrm{abc}}=100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c} ; \overline{\mathrm{cba}}=100 \mathrm{c}+10 \mathrm{~b}+\mathrm{a} \quad \text {, then: } \\
\overline{\mathrm{abc}}-\overline{\mathrm{cba}}=(100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c})-(100 \mathrm{c}+10 \mathrm{~b}+\mathrm{a}=n(a+b+c) \Rightarrow \\
99 \times(a-c)=n(a+b+c) \Rightarrow 9 \times 11 \times(a-c)=n(a+b+c) ;
\end{array}
$$
$\therefore n$ and $(a+b+c)$ must have at least one that is a multiple of 11 and 9, and $\because a 、 b 、 c$ are all integers,
$$
\text { and } 9 \geqslant a>c \geqslant 1,0 \leqslant b \leqslant 9, \quad \therefore 3=2+1+0 \leqslant a+b+c \leqslant 9+8+7=24 \text {, }
$$
The possible values of $a+b+c$ are: $9,11,18,22$,
(1) When $a+b+c=9$, according to the characteristic of being divisible by 9, the remainder when $\overline{\mathrm{abc}}$ and $\overline{\mathrm{cba}}$ are divided by 9 is 0, which contradicts the problem statement;
(2) When $a+b+c=11$, the three-digit number $\overline{a b c}$ is: $281, 362, 371, 452, 461, 524, 542, 614, 623, 632, 641, 731, 713, 812, 821, 704, 902, 605$, after checking, only 452 meets the condition; (3) When $a+b+c=22$, the three-digit number $\mathrm{abc}$ is: $985, 976, 967, 958, 796, 895$, the number that meets the condition is 0;
(4) When $a+b+c=18$, the three-digit number $\overline{\mathrm{abc}}$ is: $954, 972, 945, 963, 981, 936, 927, 918, 891, 873, 864, 846, 837, 792, 783, 765, 695, 685, 693, 594$, after checking, the number that meets the condition is 0.

In summary, $\overline{\mathrm{abc}}=452$,
The answer is: 452.
11. (10 points) The remainder when the three-digit number $\mathrm{abc}$ is divided by the sum of its digits is 1, and the remainder when the three-digit number $\mathrm{cba}$ is divided by the sum of its digits is also 1. If different letters represent different digits, and $a>c$, then $\overline{\mathrm{abc}}=$ $\qquad$ .