Example 4 The sports meet lasted for $n$ days $(n>1)$, and $m$ medals were awarded. On the first day, 1 medal plus $\frac{1}{7}$ of the remaining $m-1$ medals were awarded, on the second day, 2 medals plus $\frac{1}{7}$ of the remaining medals were awarded, and so on, until on the $n$th day, the remaining medals were all awarded. How many days did the sports meet last? How many medals were awarded in total?
Solution: Let the number of medals remaining after $k$ days of the sports meet be $x_{k}$ $(k \in \mathbf{N}, k>1)$, then the number of medals issued on the $k$-th day is
$$
\begin{array}{l}
k+\frac{1}{7}\left(x_{k-1}-k\right)=\frac{1}{7} x_{k-1}+\frac{5}{7} k, \\
x_{k}=x_{k-1}-\left(\frac{1}{7} x_{k-1}+\frac{6}{7} k\right),
\end{array}
$$

which is $\square$
$$
\begin{aligned}
x_{k-1}= & \frac{7}{6} x_{k}+k(k \in \mathbf{N}, k>1) \\
m= & 1+\frac{7}{6} x_{1} \\
= & 1+\frac{7}{6}\left(2+\frac{7}{6} x_{2}\right) \\
= & 1+\frac{7}{6} \times 2+\left(\frac{7}{6}\right)^{2}\left(3+\frac{7}{6} x_{3}\right) \\
= & 1+2 \times \frac{7}{6}+3\left(\frac{7}{6}\right)^{2}+\left(\frac{7}{6}\right)^{3} x_{3} \\
= & \cdots \cdots \\
= & 1+2 \times \frac{7}{6}+3\left(\frac{7}{6}\right)^{2}+\cdots \\
& +n\left(\frac{7}{6}\right)^{n-1}+\left(\frac{7}{6}\right)^{n} x_{n}
\end{aligned}
$$

Given $x_{n}=0$, we have
$$
\begin{aligned}
m & =1+2 \times \frac{7}{6}+3\left(\frac{7}{6}\right)^{2}+\cdots+n\left(\frac{7}{6}\right)^{n-1} \\
& =\frac{(n-6) 7^{n}}{6^{n-1}}+36 .
\end{aligned}
$$

Since $n>1,|n-6|<6^{n-1}, 7^{n}$ and $6^{n-1}$ are coprime, hence $n=6, m=36$.
The sports meet lasted 6 days, and a total of 36 medals were issued.