Solution:

We immediately exclude the case in which $n$ has all digits different from zero, since in this case we have $n_{0}=n$. There remain therefore two possibilities:

1) $n$ has the third digit equal to zero; the decimal representation of $n$ is of the type $n=ab0$, $n_{0}=ab$ and $n_{0} \mid n$ for every choice of $a \in \{1, \ldots, 9\}$ and of $b \in \{0,1, \ldots, 9\}$. Regarding this case, there are thus $9 \times 10 = 90$ possibilities for $n$.

2) $n$ has the third digit different from zero but the second equal to zero; in this case $n$ is written in the form $a0b$ and $n_{0}$ in the form $ab$. Recalling the meaning of decimal notation, we thus look for the cases in which $10a + b$ divides $100a + b$. Since certainly $10a + b$ divides $10(10a + b) = 100a + 10b$, it is also equivalent to see when $10a + b$ divides the difference $(100a + 10b) - (10a + b) = 9b$.

Writing $k(10a + b) = 100a + b$ and considering the last decimal digit, we obtain that $(k-1)b$ is divisible by $10$, so that at least one of $k-1$ and $b$ is divisible by $5$. Since evidently $1 < k < 10$ and $1 \leq b \leq 9$, we have that $k-1 = 5$ or $b = 5$.

For $k-1 = 5$ we obtain $60a + 6b = 100a + b$, from which $b = 8a$ and hence $a = 1$, $b = 8$. One checks that $n = 108$ is indeed a solution, since $18$ divides $108$.

For $b = 5$ we use the equivalent property that $10a + b$ divides $9b$, which becomes $10a + 5$ divides $45$; one sees immediately that the divisors of $45$ of the form $10a + 5$ with $a \geq 1$ are $15$ and $45$, giving the two solutions $n = 105$ ($15$ divides $105$) and $n = 405$ ($45$ divides $405$).

In conclusion, there are $90$ solutions in case 1) and $3$ solutions in case 2), for a total of $93$ solutions.