Three, (25 points) Find all positive integers such that it is 224 times the sum of its digits.



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Let the required positive integer be $N$, where $N$ is an $n$-digit number, and the sum of its digits is $S$.
Then $N \geqslant 10^{n-1}, S \leqslant 9 n$.
By the problem statement,
$$
N=224 S \Rightarrow 10^{n-1} \leqslant 224 \times 9 n \text {. }
$$

It can be proven that: when $n>6$, $10^{n-1}>224 \times 9 n$.
When $n=5$, $N \leqslant 224 \times 9 \times 5=10080$.
Let $N=\overline{100 a b}$. Then
$$
\begin{array}{l}
10000+10 a+b=224(1+a+b) \\
\Rightarrow a=45-b+\frac{146-9 b}{214} .
\end{array}
$$

When $b$ is $0 \sim 9$, $\frac{146-9 b}{214}$ is not an integer.
When $n=4$, let $N=\overline{a b c d}(a \neq 0)$, then $1000 a+100 b+10 c+d=224(a+b+c+d)$. Thus, $d$ is even,
$$
\begin{array}{l}
4|(10 c+d), 4|(2 c+d) . \\
\text { Let } d=2 d_{1}\left(0 \leqslant d_{1} \leqslant 4\right),
\end{array}
$$
$$
2 c+d_{1}=4 k, c=2 k-d_{1}(0 \leqslant k<7) .
$$

Then $388 a=62 b+107 c+223 d_{1}$
$$
\begin{array}{l}
\Rightarrow b=6 a-4 k-2 d_{1}+\frac{8 a+17 k+4 d_{1}}{31} . \\
\text { Let } 8 a+17 k+4 d_{1}=31 t\left(t \in \mathbf{Z}_{+}\right) . \\
\text {Also } 8 \leqslant 8 a+17 k+4 d_{1} \\
\leqslant 8 \times 9+17 \times 6+4 \times 4=190,
\end{array}
$$

Thus $t=1,2, \cdots, 6$.
If $t=1$, then $8 a+17 k+4 d_{1}=31<34$.
Obviously, $k=1,8 a+4 d_{1}=14$, which is impossible.
If $t=2$, then $8 a+17 k+4 d_{1}=62<68$.
Obviously, $k=2,2 a+d_{1}=7$.
So, $a=3, d_{1}=1$ or $a=2, d_{1}=3$.
When $a=3, d_{1}=1, k=2$, from equation (1) we get $b=10$ (discard);
When $a=2, d_{1}=3, k=2$, from equation (1) we get $b=0, c=1, d=6$.
Therefore, 2016 meets the conditions.
Similarly, we can discuss $t=3,4,5,6$, and there are no positive integers that meet the conditions.
Thus, only 2016 is the positive integer that meets the conditions.