10.8. Given a natural number $n \geqslant 2$. Consider all colorings of the cells of an $n \times n$ board in $k$ colors such that each cell is painted in exactly one color, and all $k$ colors are used. For what smallest $k$ will any such coloring contain four cells painted in four different colors, located at the intersection of two rows and two columns?

(D. Khramov)
Answer: $k=2 n$.

First solution. First, we will show how to color the cells in $2 n-1$ colors so that there is no set of four cells of different colors lying at the intersection of two rows and two columns.

We number the horizontal rows from top to bottom with even numbers $0,2,4, \ldots, 2 n-2$, and the vertical rows from left to right with odd numbers $1,3, \ldots, 2 n-1$. We will color the 0-th row in color 0, the 1-st row in color 1, ..., and the $(2 n-1)$-th row in color $2 n-1$. In this process, when coloring each subsequent row, we repaint all the cells that were previously colored. In the final coloring, there are $2 n-1$ colors (color 0 will not remain).

Let's verify that this coloring satisfies the condition. Consider four cells at the intersection of two rows and two columns. One of these four rows (horizontal or vertical) was colored last; then the two cells in this row will remain colored in this color and, consequently, will be of the same color.

It remains to prove that for $k \geqslant 2 n$, the required set of four cells will necessarily be found. Consider the coloring of cells in $k$ or more colors. We can rephrase the problem as follows. Consider a bipartite graph where the vertices correspond to the rows and columns of the board (a total of $2 n$ vertices). If a cell at the intersection of some row and column is colored in color $c$, then we connect the vertices corresponding to these row and column with an edge of color $c$. A multicolored cycle is a cycle in which no two edges have the same color. We need to prove that in the constructed graph, there will be a multicolored cycle of 4 edges.