Let's calculate how many two-digit numbers there are in the base $n$ number system, the sum of whose digits is also two-digit (in the base $n$ system). Give the result in the base $n$ number system.
Solution. Let the two-digit numbers be plotted on a coordinate system such that the first digit of the two-digit number is the first coordinate $(x)$ of the point, while the second digit is the second coordinate $(y)$ of the point.

The first digit of a two-digit number can be anything from 1 to $(n-1)$, and the second digit can be anything from 0 to $(n-1)$, from which we immediately get that the number of two-digit numbers is $(n-1) n$.

Those numbers in which the sum of the digits is two-digit, i.e., at least $n$, are located on or above the graph of the function $y=n-x$. The largest two-digit number is $n^{2}-1$, in which the sum of the digits is $(n-1)+(n-1)=2 n-2$. It can be seen from a simple transformation that its value is at most $n^{2}-1$ (since $n^{2}-1-(2 n-2)=n^{2}-2 n+1=(n-1)^{2} \geq 0$), so it is certainly not a three-digit number.

The number of points that are on or above the line $y=n-x$ is $1+2+\ldots+(n-2)+(n-1)$, and the number of other points is $(n-1)+(n-2)+\ldots+2+1$. Therefore, exactly half of the points meet the condition. The number of suitable numbers is thus $\frac{(n-1) n}{2}$.

The two figures illustrate the solution for $n=8$ and $n=9$.

If $n$ is odd, then $n-1$ is even, and $\frac{n(n-1)}{2}=n \cdot \frac{n-1}{2}$, so the representation in base $n$ is: $\frac{\overline{n-1}}{2} ; 0$.

If $n$ is even, then $n-2$ is also even, and $\frac{n(n-1)}{2}=n \cdot \frac{n-2}{2}+\frac{n}{2}$, so the representation is: $\frac{\overline{n-2}}{2} ; \frac{n}{2}$.

Remarks. 1. Many calculated the number of two-digit numbers this way: The smallest two-digit number is $n$, and the largest is $n^{2}-1$. The number of two-digit numbers is thus $\left(n^{2}-1\right)-n+1=n^{2}-n$.

2. The fact that exactly half of the two-digit numbers are good can also be justified this way: The figures are symmetric about the point $P\left(\frac{n}{2} ; \frac{n-1}{2}\right)$, and the $n(n-1)$ two-digit numbers can be paired according to the symmetry such that one member of each pair is two-digit, and the other is not.
3. Most chose the solution path available on the website.
4. The many 3-point solutions are due to the fact that many reached the point where they found that $\frac{(n-1) n}{2}$ suitable numbers exist, but then they stopped.