Example 10 The sports meet lasted for $n(n>1)$ consecutive days, and a total of $m$ medals were awarded. On the first day, 1 medal and $\frac{1}{7}$ of the remaining $(m-1)$ medals were awarded. On the second day, 2 medals and $\frac{1}{7}$ of the remaining medals were awarded, and this pattern continued for subsequent days. On the last day, which is the $n$-th day, the remaining $n$ medals were awarded. How many days did the sports meet last, and how many medals were awarded in total?
Explanation: Suppose after $k$ days of the sports meet, there are still $a_{k}$ medals left. The number of medals issued on the $k$-th day is
$$
k+\frac{1}{7}\left(a_{k-1}-k\right),
$$

Therefore, $a_{k}=a_{k-1}-\left[k+\frac{1}{7}\left(a_{k-1}-k\right)\right]$.
Thus, $a_{k}=\frac{6}{7} a_{k-1}-\frac{6}{7} k$.
This gives us a recursive relation for the sequence $a_{k}$, with the initial and final terms being $a_{0}=m, a_{n}=0$.
Now, let's solve the recursive relation.
First, transform equation (1) into
$$
a_{k}+6 k=\frac{6}{7}\left(a_{k-1}+6 k\right),
$$

Then, transform it into
$$
a_{k}+6 k-36=\frac{6}{7}\left[a_{k-1}+6(k-1)-36\right] \text {. }
$$

This indicates that $b_{k}=a_{k}+6 k-36$ is a geometric sequence with a common ratio of $\frac{6}{7}$, and its initial and final terms are
$$
\begin{array}{l}
b_{0}=a_{0}+0-36=m-36, \\
b_{n}=a_{n}+6 n-36=6 n-36 .
\end{array}
$$

Substitute the general term formula $b_{n}=b_{0} q^{n}$, we get
$$
6 n-36=(m-36)\left(\frac{6}{7}\right)^{n} \text {, }
$$

which means $m=36+\frac{7^{n}(n-6)}{6^{n-1}}$.
Since $m$ is an integer, we have
$$
6^{n-1} \mid\left[7^{n}(n-6)\right] \text {. }
$$

However, $6^{n-1}$ and $7^{n}$ are coprime, so $6^{n-1} \mid(n-6)$.
For $n>1$, we have
$$
\begin{array}{l}
-6^{n-1}+6 \leqslant 01).
\end{array}
$$
Solving this, we get $n=6$, and thus $m=36$.
Comment: The transformation from equation (1) to equation (2), and then to equation (3), is guided by the idea of fixed points in iteration. This problem has a historical background similar to the famous "Old Man Distributing Gold Coins" problem, where exactly 6 medals are issued each day, just like the old man giving each child the same number of gold coins.