International Mathematical Olympiad

If $m=2^{a} \cdot 5^{b} \cdot s$, with $\operatorname{gcd}(s, 10)=1$, then $\frac{10^{t}-1}{m}$ is short if and only if $s$ divides $10^{t}-1$. So we may (and will) suppose without loss of generality that $\operatorname{gcd}(m, 10)=1$. Define
$$C=\{1 \leqslant c \leqslant 2017: \operatorname{gcd}(c, 10)=1\} .$$
The $m$-tastic numbers are then precisely the smallest exponents $t>0$ such that $10^{t} \equiv 1 \pmod{c m}$ for some integer $c \in C$, that is, the set of orders of $10$ modulo $c m$. In other words,
$$S(m)=\left\{\operatorname{ord}_{c m}(10): c \in C\right\} .$$
Since there are $4 \cdot 201+3=807$ numbers $c$ with $1 \leqslant c \leqslant 2017$ and $\operatorname{gcd}(c, 10)=1$, namely those such that $c \equiv 1,3,7,9 \pmod{10}$,
$$|S(m)| \leqslant |C|=807 .$$
Now we find $m$ such that $|S(m)|=807$. Let
$$P=\{1<p \leqslant 2017: p \text{ is prime, } p \neq 2,5\}$$
and choose a positive integer $\alpha$ such that every $p \in P$ divides $10^{\alpha}-1$ (e.g. $\alpha=\varphi(T)$, $T$ being the product of all primes in $P$), and let $m=10^{\alpha}-1$.

Claim. For every $c \in C$, we have
$$\operatorname{ord}_{c m}(10)=c \alpha$$
As an immediate consequence, this implies $|S(m)|=|C|=807$, finishing the problem.

Proof. Obviously $\operatorname{ord}_{m}(10)=\alpha$. Let $t=\operatorname{ord}_{c m}(10)$. Then
$$c m \mid 10^{t}-1 \quad \Longrightarrow \quad m \mid 10^{t}-1 \quad \Longrightarrow \quad \alpha \mid t .$$
Hence $t=k \alpha$ for some $k \in \mathbb{Z}_{>0}$. We will show that $k=c$.

Denote by $\nu_{p}(n)$ the number of prime factors $p$ in $n$, that is, the maximum exponent $\beta$ for which $p^{\beta} \mid n$. For every $\ell \geqslant 1$ and $p \in P$, the Lifting the Exponent Lemma provides
$$\nu_{p}\left(10^{\ell \alpha}-1\right)=\nu_{p}\left(\left(10^{\alpha}\right)^{\ell}-1\right)=\nu_{p}\left(10^{\alpha}-1\right)+\nu_{p}(\ell)=\nu_{p}(m)+\nu_{p}(\ell)$$
So
$$\begin{aligned} c m \mid 10^{k \alpha}-1 & \Longleftrightarrow \forall p \in P ; \quad \nu_{p}(c m) \leqslant \nu_{p}\left(10^{k \alpha}-1\right) \\ & \Longleftrightarrow \forall p \in P ; \quad \nu_{p}(m)+\nu_{p}(c) \leqslant \nu_{p}(m)+\nu_{p}(k) \\ & \Longleftrightarrow \forall p \in P ; \quad \nu_{p}(c) \leqslant \nu_{p}(k) \\ & \Longleftrightarrow c \mid k . \end{aligned}$$
The first such $k$ is $k=c$, so $\operatorname{ord}_{c m}(10)=c \alpha$.