Japan Junior Mathematical Olympiad

A positive integer $n$ less than or equal to $999$ can be represented as $10^2a + 10b + c$ where $a, b, c$ are integers satisfying $0 \le a, b, c \le 9$ and $a + b + c \ne 0$. Then we have $S(n) = a + b + c$ and
$$S(n+1) = \begin{cases} a + b + c + 1, & \text{if } c < 9 \cdots (\text{i}), \\ a + b + 1, & \text{if } c = 9, b < 9 \cdots (\text{ii}), \\ a + 1, & \text{if } c = b = 9, a < 9 \cdots (\text{iii}), \\ 1, & \text{if } c = b = a = 9 \cdots (\text{iv}). \end{cases}$$

Now, it is clear that $\frac{S(n)}{S(n+1)}$ is not an integer in case (i), and is the integer $999$ in case (iv) above. In case (iii), we have $\frac{S(n)}{S(n+1)} = \frac{a + 18}{a + 1}$, which is an integer only when $a = 0$, corresponding to the case $n = 99$. Finally, in case (ii), $\frac{S(n)}{S(n+1)} = \frac{a + b + 9}{a + b + 1}$, with $0 \le a \le 9$, $0 \le b < 9$. If we let $k = a + b$, then we see that $\frac{k + 9}{k + 1}$ is an integer only when $k = 0, 1, 3, 7$, from which we conclude that only possibilities for $(a, b)$ under the case (ii) are
$$(a, b) = (0, 0), (0, 1), (0, 3), (0, 7), (1, 0), (1, 2), (1, 6), (2, 1), (2, 5), (3, 0), (3, 4), (4, 3), (5, 2), (6, 1), (7, 0),$$
and these choices of $(a, b)$ correspond to
$$n = 9, 19, 39, 79, 109, 129, 169, 219, 259, 309, 349, 439, 529, 619, 709,$$
respectively. These, together with the numbers $99$ and $999$ found above, give $17$ integers satisfying the condition of the problem.