25. Among the integers between 100 and 999, there are ( ) numbers that have the property: the digits of the number can be rearranged to form a number that is a multiple of 11 and is between 100 and 999 (for example, 121 and 211 both have this property).
(A) 226
(B) 243
( C) 270
(D) 469
(E) 486
25. A.

Let a three-digit number be $\overline{A C B}$. Then
11. $\overline{A C B} \Leftrightarrow 11 \mathrm{I}(A+B-C)$
$\Leftrightarrow A+B=C$ or $A+B=C+11$.
We will discuss the following scenarios.
Note that, $A$ and $B$ are of equal status, so we can assume $A \geqslant B$ (the case for $A < B$ is similar).
(1) $A+B=C$.
If $B=0, A=C$, we get the numbers satisfying the property 110, $220, \cdots, 990$. Each number has two permutations. Therefore, there are $2 \times 9=18$ such numbers.

If $B=1, A+1=C$, we get the numbers satisfying the property $121,231, \cdots, 891$, where the first number has three permutations, and the other seven numbers each have six permutations. Therefore, there are $3+7 \times 6=45$ such numbers.

If $B=2, A+2=C$, we get the numbers satisfying the property $242,352, \cdots, 792$, where the first number has three permutations, and the other five numbers each have six permutations. Therefore, there are $3+5 \times 6=33$ such numbers.

If $B=3, A+3=C$, we get the numbers satisfying the property $363,473, \cdots, 693$, where the first number has three permutations, and the other three numbers each have six permutations. Therefore, there are $3+3 \times 6=21$ such numbers.
If $B=4, A+4=C$, we get the numbers satisfying the property
484, 594. The first number has three permutations, and the second number has six permutations. Therefore, there are 9 such numbers.
In total, there are $18+45+33+21+9=126$ numbers.
(2) $A+B=C+11$.
If $C=0, A+B=11$, we get the numbers satisfying the property $902, 803, 704, 605$. Each number has four permutations. Therefore, there are $4 \times 4=16$ such numbers.

If $C=1, A+B=12$, we get the numbers satisfying the property $913, 814, 715, 616$, where the first three numbers each have six permutations, and the last number has three permutations. Therefore, there are $3 \times 6+3=21$ such numbers.

If $C=2, A+B=13$, we get the numbers satisfying the property $924, 825, 726$. Each number has six permutations. Therefore, there are $3 \times 6=18$ such numbers.

If $C=3, A+B=14$, we get the numbers satisfying the property $935, 836, 737$. The first two numbers each have six permutations, and the last number has three permutations. Therefore, there are $2 \times 6+3=15$ such numbers.

If $C=4, A+B=15$, we get the numbers satisfying the property 946, 847. Each number has six permutations. Therefore, there are $2 \times 6=12$ such numbers.

If $C=5, A+B=16$, we get the numbers satisfying the property $957, 858$. The first number has six permutations, and the second number has three permutations. Therefore, there are 9 such numbers.

If $C=6, A+B=17$, we get the number 968. It has six permutations. Therefore, there are 6 such numbers.

If $C=7, A+B=18$, we get the number 979. It has three permutations. Therefore, there are 3 such numbers.
In total, there are
$$
16+21+18+15+12+9+6+3=100
$$

numbers.
Combining (1) and (2), the total number is
$126+100=226$.