6.26 The sports meet lasted for $n$ days, and a total of $m$ medals were awarded. On the first day, one medal and $\frac{1}{7}$ of the remaining $m-1$ medals were awarded. On the second day, two medals and $\frac{1}{7}$ of the remaining medals were awarded. This pattern continued, and on the last day, which was the $n$-th day, the remaining $n$ medals were awarded. Find the values of $n$ and $m$.
[Solution] Let the number of days the sports meet has been held be $k$, and the remaining number of medals be $a_{k}$. According to the given information, on the $k$-th day, a total of $k$ medals plus $\frac{1}{7}$ of the remaining medals are awarded. In other words, the number of medals awarded on the $k$-th day is equal to $k+\frac{1}{6} a_{k}$, hence we have
$$
a_{k-1}=k+\frac{1}{6} a_{k}+a_{k}=k+\frac{7}{6} a_{k} .
$$

Noting that $a_{n}=0$, by recursive application of (1), we get
$$
m=1+2 \times \frac{7}{6}+3 \times\left(\frac{7}{6}\right)^{2}+\cdots+n \times\left(\frac{7}{6}\right)^{n-1} .
$$

Multiplying both sides of (2) by $\frac{7}{6}$, we obtain
$$
\frac{7}{6} m=\frac{7}{6}+2 \times\left(\frac{7}{6}\right)^{2}+3 \times\left(\frac{7}{6}\right)^{3}+\cdots+n \times\left(\frac{7}{6}\right)^{n} .
$$
Subtracting (2) from (3) and simplifying, we finally get
$$
m=\frac{7^{n}(n-6)}{6^{n-1}}+36 .
$$

Since $n>1$, we have $|n-6|<6^{n-1}$. Also, since $\left(7^{n}, 6^{n-1}\right)=1$ and $m$ is a positive integer, it must be that $n-6=0$, i.e., $n=6, m=36$.