26*. The Spartakiad lasted $n$ days; during which $N$ sets of medals were awarded; on the 1st day, 1 set of medals and $1 / 7$ of the remaining quantity were awarded; on the 2nd day - 2 sets of medals and $1 / 7$ of the remaining quantity; ...; on the penultimate, $(n-1)$-th day - $(n-1)$ sets of medals and $1 / 7$ of all the remaining medals; finally, on the last day, the last $n$ sets of medals were awarded. How many days did the Spartakiad last and how many sets of medals were awarded?
26. Let the number of medal sets remaining to be awarded at the beginning of the $i$-th day of the competition be denoted by $N_{i}$, where $i=1,2, \ldots, \ldots, n$; it can, however, be assumed that the quantity $N_{i}$ makes sense even for $i>n$, becoming 0 in these cases. From the conditions of the problem (and our agreement), it follows that $N_{1}=N, N_{n}=n, N_{n+t} \equiv 0$; moreover, the values of $N_{i}$ and $N_{i+1}$ are related by the relation:

$$
\begin{aligned}
& N_{i+1}=N_{i}-i-\frac{1}{7}\left(N_{i}-i\right)=\frac{6}{7}\left(N_{i}-l\right) \\
& \quad \text { or } N_{i}=\frac{7}{6} N_{i+1}+i
\end{aligned}
$$

From $\left(^{*}\right)$ we sequentially find

$$
\begin{aligned}
& N_{n}=n\left(=\frac{7}{6} N_{n+1}+n\right) i \\
& N_{n-1}\left.=\frac{7}{6} n+(n-1)\right) \\
& N_{n-2}=\left(\frac{7}{6}\right)^{2} n+\frac{7}{6}(n-1)+(n-2) \mathbf{1} \\
& N_{n-3}=\left(\frac{7}{6}\right)^{3} n+\left(\frac{7}{6}\right)^{2}(n-1)+\frac{7}{6}(n-2)+(n-8) 1 \\
& \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
& N_{i}=\left(\frac{7}{6}\right)^{n-i} n+\left(\frac{7}{6}\right)^{n-i-1}(n-1)+\left(\frac{7}{6}\right)^{n-1-2}(n-2)+\ldots+i
\end{aligned}
$$

(The general formula, of course, can be easily proved by mathematical induction).

Thus, we obtain:

$$
\begin{aligned}
& N=N_{1}=\left(\frac{7}{6}\right)^{n-1} n+\left(\frac{7}{6}\right)^{n-2}(n-1)+\left(\frac{7}{8}\right)^{n-8}(n-2)+\ldots \\
& \quad \ldots+1=n\left[\left(\frac{7}{6}\right)^{n-1}+\left(\frac{7}{6}\right)^{n-2}+\left(\frac{7}{6}\right)^{n-3}+\ldots+1\right]- \\
& -\left[\left(\frac{7}{6}\right)^{n-2}+2\left(\frac{7}{6}\right)^{n-3}+\ldots+(n-2) \frac{7}{6}+(n-1)\right]=S_{1} \cdot n-S_{2}
\end{aligned}
$$

where $S_{1}$ and $S_{2}$ denote the two sums enclosed in square brackets. But by the formula for the sum of the terms of a geometric progression, it is clear that

$$
S_{1}=\left(\frac{7}{6}\right)^{n-1}+\left(\frac{7}{6}\right)^{n-2}+\ldots+1=\frac{\left(\frac{7}{6}\right)^{n}-1}{\frac{7}{6}-1}=6\left[\left(\frac{7}{6}\right)^{n}-1\right]
$$

On the other hand, it is easy to see that

$$
\begin{aligned}
\left(\frac{7}{6}\right) \cdot S_{2}-S_{2}=\left(\frac{7}{6}\right)^{n-1}+\left(\frac{7}{6}\right)^{n-2} & +\ldots+\left(\frac{7}{6}\right)-(n-1)= \\
& =S_{1}-n=6\left[\left(\frac{7}{6}\right)^{n}-1\right]-n
\end{aligned}
$$

so that

$$
S_{2}=36\left[\left(\frac{7}{6}\right)^{n}-1\right]-6 n
$$

and, therefore,

$$
N=6\left[\left(\frac{7}{6}\right)^{n}-1\right] n-36\left[\left(\frac{7}{6}\right)^{n}-1\right]+6 n=6(n-6)\left(\frac{7}{6}\right)^{n}+36
$$

Since $N$ is an integer, the number $\frac{7 n(n-6)}{6^{n-1}}$, and hence $\frac{n-6}{6^{n-1}}$ must also be integers; therefore, $n$ must be a multiple of 6. On the other hand, it is clear that for all $k \geqslant 2$ the inequality $6 k-66$ cannot be an integer. Thus, we arrive at the only possible solution to the problem: $n=6$ and, therefore (by (**)) $N=36$.