There is an integer $n>1$. There are $n^2$ stations on a slope of a mountain, all at different altitudes. Each of two cable car companies, $A$ and $B$, operates $k$ cable cars; each cable car provides a transfer from one of the stations to a higher one (with no intermediate stops). The $k$ cable cars of $A$ have $k$ different starting points and $k$ different finishing points, and a cable car that starts higher also finishes higher. The same conditions hold for $B$. We say that two stations are linked by a company if one can start from the lower station and reach the higher station by using one or more cars of that company (no other movements between stations are allowed).

Determine the smallest positive integer $k$ for which one can guarantee that there are two stations that are linked by both companies.

Step 1 (Graph model): Model each company's cable cars as a directed graph on $n^2$ vertices where edges go from lower to higher stations. Company A's graph has $k$ edges with the property that starting points are distinct, finishing points are distinct, and the edge function is order-preserving. This means A's edges form a partial permutation (matching) that is order-preserving — i.e., a set of $k$ non-crossing edges.

Step 2 (Connected components): The graph of company A has $n^2$ vertices and $k$ edges, giving $n^2-k$ connected components (each component is a directed path). For $k=n^2-n+1$, the number of components is $n^2-(n^2-n+1)=n-1$.

Step 3 (Pigeonhole argument): Both A and B have at most $n-1$ path components. A is linked if and only if they're in the same path component. By Dilworth's theorem or a direct pigeonhole argument: if A has $\le n-1$ components and B has $\le n-1$ components, then the number of pairs $(C_A,C_B)$ of (component of A, component of B) is at most $(n-1{)}^2$. Since there are $n^2$ stations, by pigeonhole, two stations share both the same A-component and same B-component if $n^2>(n-1{)}^2$, i.e., $2n-1>0$, which is always true.

Step 4 (Construction for $k=n^2-n$): With $k=n^2-n$ edges, each company has $n$ components. Arrange the $n^2$ stations in an $n\times n$ grid. Company A's components are the rows, company B's components are the columns. Then no two stations share both the same row and column component (each cell is unique). This shows $k=n^2-n$ is insufficient.

Conclusion: $k=n^2-n+1$. $■$