The last digit is not 0, after removing one digit (but not the first digit) from it, it becomes a smaller number that is an integer multiple of the resulting number (i.e., the original number is an integer multiple of the resulting number -- translator's note).





Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
57. 19. Let the original number be $b$ times the later number, i.e.,
$$
\begin{array}{l}
\overline{a_{k} \cdots a_{l+1}} \cdot 10^{t}+a_{t} \cdot 10^{t-1}+\overline{a_{i-1} \cdots a_{1}} \\
=b\left(\overline{a_{k} \cdots a_{l+1}} \cdot 10^{t-1}+\overline{a_{t-1} \cdots a_{1}}\right) .
\end{array}
$$

It is easy to see that $110$, then $l=k-1$, and from (1) we get
$$
\begin{array}{l}
{\left[a_{k-1}-(b-10) a_{k}\right] \cdot 10^{4-2}} \\
=(b-1) \cdot \overline{a_{k-2} \cdots a_{k}} .
\end{array}
$$
$$
u_{k-1}>(b-10) a_{k} \text {. }
$$

At this point, (2) becomes
$$
\begin{array}{l}
(b-1) \cdot \overline{a_{k-2} \cdots a_{1}} \\
=2^{k-2} \cdot 5^{k-2} \cdot\left[a_{k-1}-(b-10) a_{k}\right] .
\end{array}
$$

Since $a_{1} \neq 0$, $\overline{a_{1-2} \cdots a_{1}}$ cannot be both a multiple of 2 and a multiple of 5. When it is a multiple of 2, $b-1$ should be a multiple of $5^{1-2}$. But since $b-17 a_{6} .
$$

Since $a_{6} \geqslant 1$, the only suitable values for $a_{5}$ and $a_{6}$ in (5) are:
$a_{5}=8, a_{6}=1$ or $a_{5}=9, a_{6}=1$.
Substituting $a_{5}=8, a_{6}=1, b-1=16$ into (4), we get
$$
\overline{a_{4} a_{3} a_{2} a_{1}}=\frac{10000}{16}=625 .
$$

This indicates that the original number is
$$
\overline{a_{6} a_{5} a_{4} a_{3} a_{2} a_{1}}=180625 \text {. }
$$

In fact, $180625=17 \times 10625$. This meets the requirement.
Substituting $a_{5}=9, a_{6}=1, b-1=16$ into (4), we get
$$
\overline{a_{4} a_{3} a_{2} a_{1}}=\frac{20000}{16}=1250 .
$$

Since $a_{i}=0$ in this case, it does not meet the requirement.
Next, we discuss $b1$. After similar discussions, we find that when $b-1=$ $8, l-1=3, \overline{a_{3} a_{2} a_{1}}=875$, the right side of (7) is maximized, and we have
$$
\overline{a_{4} \cdots a_{5}}+a_{4}=7 .
$$

Thus, $k=5, a_{5} \leqslant 7$. Therefore, the largest positive integer in this case is 70875.
Since $70875=9 \times 7875$. And $70875<180625$, the largest positive integer that meets the requirement is 180625.