Initially 212 It is known that there is a four-digit number, and the sum of this four-digit number and the sum of its digits is 2008. Find this four-digit number and explain the reason.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Solution: Let the four-digit number be $\overline{a b c d}$. According to the problem,
$$
1000 a+100 b+10 c+d+a+b+c+d=2008 \text {, }
$$

which simplifies to $1001 a+101 b+11 c+2 d=2008$.
Obviously, when $a \geqslant 3$, $1001 a>3000>2008$, so $a \geqslant 3$ is impossible. Therefore, $a=1$ or $a=2$.
(1) When $a=1$, subtract 1001 from both sides of equation (1) to get $101 b+11 c+2 d=1007$.
Since the maximum value of $11 c+2 d$ is $11 \times 9+2 \times 9=117$, then $101 b \geqslant 1007-117$. Therefore, $b>8$.
Upon inspection, only $b=9$ works.
Thus, $11 c+2 d=1007-909=98$.
Since $0 \leqslant 2 d \leqslant 18$, then $98-18 \leqslant 11 c \leqslant 98$.
Therefore, $7<c<9$. Hence, $c=8$.
When $c=8$, $11 c+2 d=11 \times 8+2 d=98$, yielding $d=5$. In this case, the four-digit number is 1985.
(2) When $a=2$, subtract 2002 from both sides of equation (1) to get $101 b+11 c+2 d=6$.
Clearly, only when $b=0, c=0, d=3$ does it hold.
In this case, the four-digit number is 2003.
In summary, the four-digit numbers sought are 1985 or 2003.
(Zhu Shengkun, Liangtinghe Middle School, Luotian County, Hubei Province, 438626)