$4 \cdot 215$ The sports meet lasted for $n$ days, and a total of $m$ medals were issued. On the first day, 1 medal was issued, as well as $\frac{1}{7}$ of the remaining $(m-1)$ medals. On the second day, 2 medals were issued, as well as $\frac{1}{7}$ of the remaining medals after that. This pattern continued for subsequent days, and on the last day, which is the $n$-th day, the remaining $n$ medals were issued. How many days did the sports meet last? How many medals were issued in total?
【Solution】Let the sports meet last for $k$ days, and there are still $a_{k}$ medals left. Where $k=0,1$, $2, \cdots, n$, and $a_{0}=m, a_{n}=0$. Then the number of medals issued on the $k$-th day is:
$$
k+\frac{1}{7}\left(a_{k-1}-k\right)=\frac{1}{7} a_{k-1}+\frac{6}{7} k .
$$

So
$$
\begin{aligned}
a_{k} & =a_{k-1}-\left(\frac{1}{7} a_{k-1}+\frac{6}{7} k\right) \\
& =a_{k-1}-\frac{1}{7} a_{k-1}-\frac{6}{7} k \\
& =\frac{6}{7} a_{k-1}-\frac{6}{7} k,
\end{aligned}
$$

i.e., $a_{k-1}=k+\frac{7}{6} a_{k}$.
Thus, from this recurrence relation, we get
$$
\begin{aligned}
m=1+ & \frac{7}{6} a_{1} \\
=1+ & \frac{7}{6}\left(2+\frac{7}{6} a_{2}\right) \\
= & 1+2 \cdot \frac{7}{6}+\left(\frac{7}{6}\right)^{2} a_{2} \\
= & 1+2 \cdot \frac{7}{6}+\left(\frac{7}{6}\right)^{2}\left(3+\frac{7}{6} a_{3}\right) \\
= & 1+2 \cdot \frac{7}{6}+3 \cdot\left(\frac{7}{6}\right)^{2}+\left(\frac{7}{6}\right)^{3} a_{3} \\
& \cdots \cdots \\
= & 1+2 \cdot \frac{7}{6}+3 \cdot\left(\frac{7}{6}\right)^{2}+\cdots+n \cdot\left(\frac{7}{6}\right)^{n-1}+\left(\frac{7}{6}\right)^{n} . \\
& a_{n} .
\end{aligned}
$$

Since $a_{n}=0$, we have
$$
\begin{array}{l}
m=1+2 \cdot \frac{7}{6}+3 \cdot\left(\frac{7}{6}\right)^{2}+\cdots+n \cdot\left(\frac{7}{6}\right)^{n-1}, \\
\frac{7}{6} m=\frac{7}{6}+2 \cdot\left(\frac{7}{6}\right)^{2}+3 \cdot\left(\frac{7}{6}\right)^{3}+\cdots+n \cdot\left(\frac{7}{6}\right)^{n} .
\end{array}
$$

Subtracting the two equations, we get
$$
m-\frac{7}{6} m=1+\frac{7}{6}+\left(\frac{7}{6}\right)^{2}+\cdots+\left(\frac{7}{6}\right)^{n-1}-n \cdot\left(\frac{7}{6}\right)^{n} .
$$

That is, $-\frac{1}{6} m=\frac{\left(\frac{7}{6}\right)^{n}-1}{\frac{7}{6}-1}-n \cdot\left(\frac{7}{6}\right)^{n}$,
so $m=\frac{7^{n}(n-6)}{6^{n-1}}+36$.
For $n>1$, it is clear that
$$
|n-6|<6^{n-1},\left(7^{n}, 6^{n-1}\right)=1 .
$$

Since $m$ is an integer, it must be that $n-6=0$.
That is, $n=6, m=36$.
This means the sports meet lasted for 6 days, and 36 medals were issued.