If there are $n$ stations on the line, each station must have $n-1$ sets of tickets; in total, $n(n-1)$ sets of tickets are needed. If there were $x$ stations on the line so far and it is planned to have $y$ stations in the future, $y(y-1)-x(x-1)$ new sets of tickets will be needed.

$$
y(y-1)-x(x-1)=46 \rightarrow y^{2}-x^{2}-(y-x)=46 \rightarrow(y-x)(y+x-1)=46
$$

Both factors must be whole and positive numbers; this results in only two possibilities: $46=2 \cdot 23 ; 46=1 \cdot 46$. In the first case, $y-x=2$ and $y+x-1=23$. The equations yield $x=11, y=13$. Therefore, there were 11 stations on the line so far, and after the opening of two new stations, there will be 13.

The second case is invalid, as it would only involve opening one new station, while the problem statement refers to "stations" in the plural.
## 76. Math Puzzle $9 / 71$

Along a railway line, new passenger stations were added. To ensure that tickets for every station on the line are available at every other station, 46 additional ticket sets (new tickets between two stations) were printed before the opening of the new stations.

How many stations were already on the line, and how many new stations were built?