Answer: $1$.
Let's label the columns of the table from bottom to top with the numbers from $1$ to $2022$ and color all the cells of the rows with odd numbers in yellow, and with even numbers — in blue. Each $L$-tetromino consists of three squares of one color and one square of another. Let's assume that not a single $Z$-tetromino has been obtained. Since the numbers of yellow and blue cells in the table are equal, the number of $L$-tetrominoes consisting of three yellow cells and one blue cell is equal to the number of $L$-tetrominoes consisting of three blue cells and one yellow cell. This means that the board is divided into an even number of $L$-tetrominoes, but this implies that the total number of cells on the board is a multiple of $8$, which is not true. Hence at least one $Z$-tetromino has been obtained.
Let's give an example showing what exactly one $Z$-tetromino can be obtained.

The picture shows how to cut a $6 \times 6$ board into seven $L$-tetrominoes and one $Z$-tetromino. If we cut off such $6 \times 6$ board from the corner of the $2022 \times 2022$ board, then the rest of the board can be easily cut into $2 \times 4$ and $4 \times 2$ rectangles, which are made up of two $L$-tetrominoes.