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
$126+100=226$.