17. Let $n$ be the number of cities currently served. The corresponding number of flights is $n(n-1)$, since each city is connected to $n-1$ cities served by the airline. For the next year, the airline plans

$$
(n+k)(n+k-1)
$$

flights, where $k$ is the additional number of cities. Therefore,

$$
(n+k)(n+k-1)-n(n-1)=76
$$

From this,

$$
k(k+2 n-1)=76(=4 \cdot 19=2 \cdot 39=1 \cdot 76)
$$

Thus, $k=4$, or $k=2$, or, finally, $k=1$.

But if $k=2$, then $k+2 n-1=38$, i.e., $2 n-1=36$, which is impossible since $(2 n-1)$ is an odd number. Therefore, the only possible cases are: $k=4$ - and then $k+2 n-1=19$, i.e., $n=8$, and $k=1$ - and then $k+2 n-1=76$, i.e., $n=38$.

Thus, either the airline currently serves 8 cities, and next year it will serve 12 cities, or it currently serves 38 cities, and next year the number will increase to 39.
## 17. Airline

One American airline serves several major cities, establishing direct flights between each pair of these cities. Next year, it plans to increase the number of flights by 76, which will allow it to serve some additional cities under the same conditions.

How many cities does this airline currently serve, and how many cities does it plan to serve next year?