Indeed, let's prove that a Fantastic number can have at most $7$ digits (perhaps this is not sharp, but certainly good enough to solve the problem). Let's suppose that $n$ has more than $7$ digits and is fantastic. We distinguish cases on which digit we remove. First case: We remove the last digit: So write $n=10a+b$, then $a \mid n$ and hence $a \mid b$ and hence $a \le 9$ (since $b \ne 0$) and hence $n<100$ contradicting our assumption. Second case: We remove a digit which is neither the last nor the first nor the second. So write $n=10^{k+1}a+10^kb+c$ with $a>10$ and where $0<c<10^k$ and $1 \le b \le 9$ is the digit we remove. So $10^ka+c \mid n$ but it also divides $10^{k+1}a+10c$, hence $10^ka+c \mid 10^kb-9c$. But the RHS is not a multiple of $10$, hence in particular non-zero. On the other hand, its absolute value is clearly less than $10^{k+1}$ while $10^ka+c>10^{k+1}$. Contradiction! Third case: We remove the first digit. So write $n=10^kb+c$ where $0<c<10^k$ and $1 \le b \le 9$ is the digit we remove and $k \ge 7$. So $c \mid 10^kb+c$ and hence $c \mid 10^kb$. But $c$ cannot be divisible by $10$. If $c$ is even, then it is coprime to $5$, hence $c \mid 2^kb$. If $c$ is odd, then $c \mid 5^kb$. In any case, $c \le 5^kb$. On the other hand, the first digit of $c$ does not vanish, so $c \ge 10^{k-1}$ and hence $2^k \le 10b \le 90$ which is a contradiction to $k \ge 7$. Fourth case: We remove the second digit. As in the second case, we write $n=10^{k+1}a+10^kb+c$ but now $1 \le a \le 9$. But as before, we find that $10^ka+c \mid 10^kb-9c$ and hence also $10^ka+c \mid 10^k(9a+b)$. But then $10^ka+c \le 5^k(9a+b) \le 90 \cdot 5^k$ as in the second case and hence $2^k \le 90$ contradicting $k \ge 7$.