Example 6 A sports meet lasted for $n(n>1)$ days, and a total of $m$ medals were awarded. On the first day, one medal was awarded, and then $\frac{1}{7}$ of the remaining $m-1$ medals were awarded. On the second day, two medals were awarded, and then $\frac{1}{7}$ of the remaining medals were awarded, and so on. Finally, on the $n$-th day, $n$ medals were awarded, and no medals were left. Question: How many days did the sports meet last? How many medals were awarded in total?
Solution 1: Let $u_{k}$ be the number of medals remaining before the $k$-th day, then the number of medals awarded on the $k$-th day is $k+\frac{1}{7}(u_{k}-k)$,
thus $\quad u_{k+1}=u_{k}-\left(k+\frac{1}{7}(u_{k}-k)\right)=\frac{6}{7}(u_{k}-k)$,
so $u_{k}-\frac{7}{6} u_{k+1}=k$.
Given $u_{1}=m, u_{n}=n$, we have
$$
\begin{array}{c}
m-\frac{7}{6} u_{2}=1, \\
u_{2}-\frac{7}{6} u_{3}=2, \\
\cdots \\
u_{n-1}-\frac{7}{6} n=n-1,
\end{array}
$$

Multiplying the $k$-th equation by $\left(\frac{7}{6}\right)^{k-1}$ and summing, we get
$$
\begin{array}{l}
m=1+2\left(\frac{7}{6}\right)+3\left(\frac{7}{6}\right)^{2}+\cdots+(n-1)\left(\frac{7}{6}\right)^{n-2}+n\left(\frac{7}{6}\right)^{n-1}, \\
\frac{7}{6} m=\frac{7}{6}+2\left(\frac{7}{6}\right)^{2}+\cdots+(n-1)\left(\frac{7}{6}\right)^{n-1}+n\left(\frac{7}{6}\right)^{n},
\end{array}
$$

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

Thus, $m=36+(n-6) \frac{7^{n}}{6^{n-1}}$.
When $n>1$, $|n-6|1$, it is clear that $\quad|n-6|<6^{n-1},\left(7^{n}, 6^{n-1}\right)=1$,
From equations (1) and (2), and given that $m$ and $n$ are positive integers, we get $n=6$, hence $m=36$.
Therefore, the sports meet lasted 6 days, a total of 36 medals were awarded, and exactly 6 medals were awarded each day on the 6th day.