The initial 267 four-digit number $w_{1}$ and the sum of its four digits is the four-digit number $w_{2}, w_{2}$ and the sum of its four digits is the four-digit number $w_{3}, w_{3}$ and the sum of its four digits is the four-digit number $w_{4}, w_{4}$ and the sum of its four digits is the four-digit number $w_{5}, w_{5}$ and the sum of its four digits is 2009. Find $w_{1}$.
Let $w_{5-i}=a_{i} b_{i} c_{i} d_{i}(i=0,1, \cdots, 4)$. Then $1001 a_{0}+101 b_{0}+11 c_{0}+2 d_{0}=2009$.
Obviously, $a_{0}=1$ (otherwise, if $a_{0}=2$, then $101 b_{0}+11 c_{0}+2 d_{0}=7$, which only has the solution $b_{0}=c_{0}=0, d_{0}=\frac{7}{2}$, which is not an integer).
Thus, $101 b_{0}+11 c_{0}+2 d_{0}=1008$.
If $b_{0} \leqslant 8$, then
$$
\begin{array}{l}
101 b_{0}+11 c_{0}+2 d_{0} \leqslant 808+99+18 \\
=925925$.
Therefore, $b_{1}=9$.
Thus, $11 c_{1}+2 d_{1}=80$, which only has the solution $c_{1}=6, d_{1}=7$.
So, $w_{4}=1967$.
Then $1001 a_{2}+101 b_{2}+11 c_{2}+2 d_{2}=1967$.
Obviously, $a_{2}=1$.
Thus, $101 b_{2}+11 c_{2}+2 d_{2}=966>925$.
Therefore, $b_{2}=9$.
Thus, $11 c_{2}+2 d_{2}=57$, which only has the solution $c_{2}=5, d_{2}=1$.
So, $w_{3}=1951$.
Then $1001 a_{3}+101 b_{3}+11 c_{3}+2 d_{3}=1951$.
Obviously, $a_{3}=1$.
Thus, $101 b_{3}+11 c_{3}+2 d_{3}=950>925$.
Therefore, $b_{3}=9$.
Thus, $11 c_{3}+2 d_{3}=41$, which only has the solution $c_{3}=3, d_{3}=4$.
So, $w_{2}=1934$.
Then $1001 a_{4}+101 b_{4}+11 c_{4}+2 d_{4}=1934$.
Obviously, $a_{4}=1$.
Thus, $101 b_{4}+11 c_{4}+2 d_{4}=933>925$.
Therefore, $b_{4}=9$.
Thus, $11 c_{4}+2 d_{4}=24$, which only has the solution $c_{4}=2, d_{4}=1$.
Therefore, $w_{1}=1921$.
(Tian Yonghai, Shao Hua Educational College, Heilongjiang Province, 152054)