Japan Mathematical Olympiad Initial Round

Let $M$ be the maximum number we seek. First we show that $M \le 939$ must hold.
Let $m, n$ be positive integers satisfying the conditions of the problem, and let $k$ be the number of digits of $n$ (so, $1 \le k \le 1000$). Let us represent $\frac{m}{n}$ as
$$\frac{m}{n} = a_1 10^{b_1} + a_2 10^{b_2} + \dots + a_N 10^{b_N} + \epsilon,$$
where $1 \le a_i \le 9$, for $i = 1, 2, \dots, N$, $0 \le b_N < b_{N-1} < \dots < b_1$, and $0 \le \epsilon < 1$. Then from $10^{999} \le m < 10^{1000}$ and $10^{k-1} \le n < 10^k$ it follows that we have $b_1 \le 1000 - k$. We can also show that the following lemma holds:

Lemma: $b_i - b_{i+1} \le k+1$ holds for every $i, 1 \le i \le N-1$, and $b_N \le k$ holds also.

Proof of Lemma: Suppose for some $i, 1 \le i \le N-1$, $b_i - b_{i+1} \ge k+2$ is satisfied. Then, if we let
$$A = a_1 10^{b_1 - b_{i+1} - k - 2} + a_2 10^{b_2 - b_{i+1} - k - 2} + \dots + a_i 10^{b_i - b_{i+1} - k - 2},$$
$$B = a_{i+1} 10^{b_{i+1}} + a_{i+2} 10^{b_{i+2}} + \dots + a_N 10^{b_N} + \epsilon,$$
we see that $A$ is a positive integer and $B$ satisfies $0 < B < 10^{b_i + 1 + k}$.
We see that with $A$ and $B$ as above we can write $\frac{m}{n} = 10^{b_i+1+k+2} A + B$. But then, since we have
$$10^{b_i+1+k+2} A_n < m = 10^{b_i+1+k+2} A_n + Bn < 10^{b_i+1+k+2} A_n + 10^{b_i+1+k+1},$$
we have to conclude that the $10^{b_i+1+k+1}$-th digit of $m$ is $0$. But this contradicts the assumption that all the digits of $m$ are not zero, and therefore, $b_i - b_{i+1} \le k+1$ must be satisfied for every $i, 1 \le i \le N-1$.

Next assume that $b_N \ge k+1$ is satisfied. If we let
$$C = a_1 10^{b_1 - k - 1} + a_2 10^{b_2 - k - 1} + \dots + a_N 10^{b_N - k - 1},$$
then we see that $C$ is a positive integer and we can write $\frac{m}{n} = 10^{k+1} C + \epsilon$. But then we have
$$10^{k+1} C_n \le m = 10^{k+1} C_n + \epsilon n < 10^{k+1} C_n + 10^k,$$
which implies that the $10^k$-th digit of $m$ is $0$, contradicting the assumption of the problem. Therefore, we must have $b_N \le k$, and this completes the proof of Lemma.

Going back to the proof of the claim that $M \le 939$, we note that the number of $0$ digits in $\left\lfloor \frac{m}{n} \right\rfloor$ equals $b_1 + 1 - N$. By summing the corresponding sides of the following $N$ inequalities shown in Lemma above
$$b_1 - b_2 \le k+1,\ b_2 - b_3 \le k+1,\ \dots,\ b_{N-1} - b_N \le k+1,\ b_N \le k,$$
we obtain $b_1 \le (k+1)N - 1$, which yields $\frac{b_1+1}{k+1} \le N$. Consequently, we get
$$\begin{aligned} b_1 + 1 - N &\le b_1 + 1 - \frac{b_1+1}{k+1} \\ &= \frac{k b_1 - 1}{k+1} + 1 \le \frac{k(1000-k)-1}{k+1} + 1 \\ &= 1003 - \left( (k+1) + \frac{1002}{k+1} \right) \le 1003 - 2\sqrt{1002} < 940, \end{aligned}$$
where we used the fact $(k+1) + \frac{1002}{k+1} \ge 2\sqrt{(k+1) \cdot \frac{1002}{k+1}} = 2\sqrt{1002}$. This shows our claim that $M \le 939$.

Finally, we show that $M \ge 939$ by exhibiting a particular choice of $m$ and $n$, satisfying the conditions of the problem, for which $M = 939$. For this purpose, consider
$$m = \overbrace{211\cdots1}^{31\text{ digits}} \overbrace{1266\cdots6}^{32\text{ digits}} \overbrace{1266\cdots6}^{32\text{ digits}} \overbrace{1266\cdots6}^{32\text{ digits}} \overbrace{1266\cdots6}^{9\text{ digits}}, \quad n = \overbrace{211\cdots1}^{31\text{ digits}}$$
Then, we see that $\left\lfloor \frac{m}{n} \right\rfloor = 1 \overbrace{00\cdots06}^{32\text{ digits}} \overbrace{00\cdots06}^{32\text{ digits}} \overbrace{00\cdots06}^{32\text{ digits}} \overbrace{00\cdots06}^{32\text{ digits}} \overbrace{00\cdots0}^{9\text{ digits}}$ and, therefore the number of its $0$-digits is $939$. Thus, we conclude that $939$ is the desired answer to the problem.