$1 \cdot 56$ Try to find the largest natural number with the following property: its last digit is not 0, and after deleting one of its digits (but not the first digit), the original number is exactly an integer multiple of the resulting number.


[Solution] Let the original number be $b$ times the obtained number, where $b$ is a positive integer. Then, we have
$$
\begin{aligned}
& \overline{a_{k} \cdots a_{l+1}} \cdot 10^{l}+a_{l} \cdot 10^{l-1}+\overline{a_{l-1} \cdots a_{1}} \\
= & b\left(\overline{a_{k} \cdots a_{l+1}} \cdot 10^{l-1}+\overline{a_{l-1} \cdots a_{1}}\right)
\end{aligned}
$$

It is easy to see that if $110$, then $k=l+1, l=k-1$. At this point, equation (1) can be simplified to
$$
a_{k} \cdot 10^{k-1}+a_{k-1} \cdot 10^{k-2}+\overline{a_{k-2} \cdots a_{1}}=b\left(a_{k} \cdot 10^{k-2}+\overline{a_{k-2} \cdots a_{1}}\right),
$$

which simplifies to
$$
\left[a_{k-1}-(b-10) a_{k}\right] \cdot 10^{k-2}=(b-1) \cdot \overline{a_{k-2} \cdots a_{1}},
$$

From the above equation, we know
$$
a_{k-1}>(b-10) a_{k},
$$

and when $\overline{a_{k-2} \cdots a_{1}}$ is a multiple of 2, noting that $a_{1} \neq 0$, we get that $b-1$ is a multiple of $5^{k-2}$. Thus, we have
$$
\begin{array}{l}
5^{k-2} \leqslant b-1 \leqslant 18, \\
k \leqslant 3,
\end{array}
$$

which means the original number in this case has at most 3 digits. Similarly, when $\overline{a_{k-2} \cdots a_{1}}$ is a multiple of 5, we get that $b-1$ is a multiple of $2^{k-2}$. Thus, we have
$$
\begin{array}{l}
2^{k-2} \leqslant b-1 \leqslant 18, \\
k \leqslant 6,
\end{array}
$$

which means the original number in this case has at most 6 digits. If $\overline{a_{k-2} \cdots a_{1}}$ is a multiple of 625, then by (3) we know $a_{5}>7 a_{6}$, so $a_{6}=1, a_{5}=8$ or 9. Substituting these results into (2) gives
$$
[8-7 \times 1] \cdot 10^{4}=16 \cdot \overline{a_{4} a_{3} a_{2} a_{1}},
$$

or
$$
[9-7 \times 1] \cdot 10^{4}=16 \cdot \overline{a_{4} a_{3} a_{2} a_{1}}.
$$

From (4), we get $\overline{a_{4} a_{3} a_{2} a_{1}}=5^{4}=625$.
At this point, the original number $\overline{a_{6} a_{5} a_{4} a_{3} a_{2} a_{1}}=180625$.
Since $180625=17 \times 10625$, 180625 is a number that meets the requirements of the problem.
From (5), we get $\overline{a_{4} a_{3} a_{2} a_{1}}=2 \times 5^{4}=1250$,
which clearly does not meet the requirement $a_{1} \neq 0$.
If $b1$.
If $\overline{a_{l-1} \cdots a_{1}}$ is a multiple of 2, from (6) we know that $b-1$ is a multiple of $5^{l-1}$. Thus, we have
$$
\begin{array}{l}
5^{l-1} \leqslant b-1 \leqslant 8, \\
l=2 .
\end{array}
$$

Substituting into (6) gives
$$
\left[(10-b) \overline{a_{k} \cdots a_{3}}+a_{2}\right] \cdot 10=(b-1) \cdot a_{1} \leqslant 72,
$$

so $k \leqslant 3$,
which means the original number has at most 3 digits.
If $\overline{a_{l-1} \cdots a_{1}}$ is a multiple of 5, then $b-1$ is a multiple of $2^{l-1}$, so we have
$$
2^{l-1} \leqslant b-1 \leqslant 8,
$$
$l \leqslant 4$.
When $b-1=8, l=4, \overline{a_{3} a_{2} a_{1}}=7 \times 5^{3}=875$, the right side of (6) is maximized, at which point (6) becomes
$$
\overline{a_{k} \cdots a_{5}}+a_{4}=7.
$$

Thus, $k=5$,
which means the original number has 5 digits.
In summary, the largest positive integer that meets the requirements of the problem is the six-digit number 180625.