If we divide a three-digit, decimal number by its reverse, the quotient is 3, and the remainder is the sum of the digits of the number. What could this number be?
Let the three-digit number be $100a + 10b + c$. According to the problem statement,

$$
100a + 10b + c = 3(100c + 10b + a) + a + b + c
$$

After rearranging and simplifying, we get

$$
32a - 100c = 7b
$$

The left side is divisible by 4, so $7b$ must also be divisible by 4, and $b$ can be at most 9. This is true in two cases, when $b=4$ or $b=8$. (The case $b=0$ is trivially not possible.)

If $b=4$, then $8a = 7 + 25c$, where $0 \leq a \leq 9$, so $7 + 25c \leq 72$, and $c \leq \frac{65}{25} < 3$.

For $c=0$, $a$ is not an integer.

For $c=1$, $8a = 32$, so $a = 4$.

For $c=2$, $8a = 57$, and $a$ is not an integer.

In this case, the only solution to the problem is $a=4, b=4, c=1$. (Indeed, $441 = 144 \cdot 3 + 9$, and $4 + 4 + 1 = 9$.)

If $b=8$, then $32a - 100c = 56$, or $8a = 25c + 14$, and now $c \leq \frac{72 - 14}{25} = \frac{58}{25}$.

In this case, $c=2$ is the only possibility, and then $a=8$; the three-digit number is 882. Therefore, the numbers that satisfy the conditions of the problem are 441 and 882.