Four, (20 points) A four-digit number, the sum of this four-digit number and the sum of its digits is 1999. Find this four-digit number and explain your reasoning.
Let this number be abcd. According to the problem, we have
$$
\begin{array}{l}
1000 a+100 b+10 c+d+a+b+c+d \\
=1999,
\end{array}
$$

which simplifies to $1001 a+101 b+11 c+2 d=1999$.
(1) Clearly, $a=1$. Otherwise, $1001 a>2000$. Subtracting $1001$ from both sides, we get
$$
101 b+11 c+2 d=998 \text{. }
$$
(2) The maximum value of $11 c + 2 d$ is $99 + 18 = 117$, so $101 b \geq 998 - 117 = 881$, which means $b=9$. Then we have
$$
11 c+2 d=998-909=89.
$$
(3) Since $0 \leq 2 d \leq 18$, it follows that $89 - 18 = 71 \leq 11 c \leq 89$. Therefore, $c=7$ or $c=8$.
When $c=7$, $11 c+2 d=77+2 d=89$, so $d=6$;
When $c=8$, $11 c+2 d=88+2 d=89$, so $d=\frac{1}{2}$. (discard).
Thus, this four-digit number is 1976.