In the unit squre For the given natural number $n \geq 2$ find the smallest number $k$ that from each set of $k$ unit squares of the $n$x$n$ chessboard one can achoose a subset such that the number of the unit squares contained in this subset an lying in a row or column of the chessboard is even
1. **Establishing the lower bound for \( k \):**
   - We need to show that \( k > 2n - 1 \). Consider the \( 2n - 1 \) squares situated on the first row and the first column of the \( n \times n \) chessboard. 
   - If we select any subset of these \( 2n - 1 \) squares, it is impossible to find a subset such that the number of squares in any row or column is even. This is because there are \( n \) squares in the first row and \( n-1 \) squares in the first column (excluding the intersection square), making a total of \( 2n - 1 \) squares. Any subset of these squares will always have an odd number of squares in at least one row or column.
   - Therefore, \( k \) must be greater than \( 2n - 1 \).

2. **Showing that \( k = 2n \) suffices:**
   - Consider a bipartite graph \( G \) with \( n \) vertices in each class. One class represents the rows of the chessboard, and the other class represents the columns.
   - Two vertices are connected by an edge if there is a square at the intersection of the corresponding row and column. Thus, the \( 2n \) squares we select correspond to \( 2n \) edges in this bipartite graph.
   - We need to show that this bipartite graph has at least one Eulerian subgraph. An Eulerian subgraph is a subgraph where every vertex has an even degree.
   - To show this, we first note that a bipartite graph with \( 2n \) edges and \( 2n \) vertices (since each class has \( n \) vertices) must contain at least one cycle. This is because a bipartite graph with \( 2n \) vertices and \( 2n \) edges cannot be a forest (a graph with no cycles) since a forest with \( 2n \) vertices can have at most \( 2n - 1 \) edges.
   - Since our graph has \( 2n \) edges, it must contain at least one cycle. From any graph that contains a cycle, we can extract an elementary cycle (a cycle with no repeated vertices except for the starting and ending vertex).
   - In this elementary cycle, each vertex has degree 2, which is even. Thus, this cycle forms an Eulerian subgraph.

3. **Conclusion:**
   - Since we have shown that \( k = 2n \) is sufficient to ensure that there is a subset of squares such that the number of squares in any row or column is even, and we have also shown that \( k > 2n - 1 \), the smallest number \( k \) that satisfies the condition is \( 2n \).

The final answer is \( \boxed{2n} \).