II. (16 points) Find all four-digit numbers that satisfy the following conditions: they are divisible by 111, and the quotient obtained is equal to the sum of the digits of the four-digit number.
$$
\overline{a b c d}=a \times 10^{3}+b \times 10^{2}+c \times 10+d
$$

satisfies the condition. Then
$$
\begin{array}{l}
\frac{a \times 10^{3}+b \times 10^{2}+c \times 10+d}{11 i} \\
=9 a+b+\frac{a-11 b+10 c+d}{11 i} . \\
\because-98 \leqslant a-11 b+10 c+d \leqslant 108, \text { and } \overline{a b c d} \text { is divisible by }
\end{array}
$$
111,
$$
\therefore a-11 b+c+d=0 \text {, }
$$

i.e., $11 b=a+10 c+d$.
According to the problem, $9 a+b=a+b+c+d$,

i.e., $8 a=c+d$.
Substituting into (1) yields
$$
11 b=9(a+c) \text {. }
$$

And from $c+d \leqslant 18$, we know $a=1$ or 2.
Thus, from equation (2) we get
$$
b=9, a=2, c=9 \text {. }
$$

Then, from $8 a=c+d$, we get $d=7$.
Therefore, the required four-digit number is 2997.