1.56 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 an integer multiple of the resulting number.

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[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 transformed into
$$a_{k} \cdot 10^{k-1}+a_{k-1} \cdot 10^{k-2}+\overline{a_{k-2} \cdots a_{1}}=b\left(a_{k} \cdot 10^{k-2}+\overline{a_{k-2} \cdots a_{1}}\right)$$

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

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,
$$\begin{array}{l}
5^{k-2} \leqslant b-1 \leqslant 18 \\
k \leqslant 3
\end{array}$$

That is, 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,
$$\begin{array}{l}
2^{k-2} \leqslant b-1 \leqslant 18 \\
k \leqslant 6
\end{array}$$

That is, the original number in this case has at most 6 digits.
If $k=6$, then $b-1$ is a multiple of $2^{4}$, thus $b=17, \overline{a_{4} a_{3} a_{2} a_{1}}$ is a multiple of $5^{4}=625$. Also, from (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,
$$\begin{array}{l}
5^{l-1} \leqslant b-1 \leqslant 8 \\
l=2
\end{array}$$

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

Therefore, $k \leqslant 3$,
that is, the number of digits of the original number is no more than 3.
If $\overline{a_{l-1} \cdots a_{1}}$ is a multiple of 5, then $b-1$ is a multiple of $2^{l-1}$, thus
$$\begin{array}{l}
2^{l-1} \leqslant b-1 \leqslant 8 \\
l \leqslant 4
\end{array}$$

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 this point, (6) becomes
$$\overline{a_{k} \cdots a_{5}}+a_{4}=7$$

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