Answer: $2020$.

Let Dima wrote down the numbers $d_1 > d_2 > \dots > d_{2022}$ and Vlad wrote down the numbers $v_1 > v_2 > \dots > v_{2022}$. Suppose the answer in the problem is $2022$ or $2021$, then $d_{2022} > v_2$. Each of the $2021$ lines not containing $v_1$ contains at least $2021$ numbers not exceeding $v_2$: the number written by Vlad and all smaller numbers. And each column contains at least $2$ numbers not less than $d_{2022}$: the number written by Dima and all greater numbers. Therefore the total amount of numbers in the table is not less than
$$2021 \cdot 2021 + 2022 \cdot 2 = 2022 \cdot 2022 + 1,$$
which exceeds the number of all numbers in the table — a contradiction.

Let's show that it could turn out that $2020$ of numbers written by Vlad are less than any number written by Dima. Fill in the table

| $v_1$ | $d_1$ | $\infty$ | $d_4$ | $d_5$ | $d_6$ | ... | $d_{2022}$ |
|-------|-------|----------|-------|-------|-------|-----|------------|
| $d_2$ | $v_2$ | $d_3$ | $\infty$ | $y$ | $y$ | ... | $y$ |
| $x$ | | $v_3$ | | | | | |
| $x$ | | | $v_4$ | | | | |
| $x$ | | | | $v_5$ | | | |
| $x$ | | | | | $v_6$ | | |
| $\vdots$ | | | | | | $\vdots$ | |
| $x$ | | | | | $v_6$ | | $v_{2022}$ |

numbers from $1$ to $2022^2$ so that
$$\infty > v_1 > v_2 > y > d_1 > \dots > d_{2022} > x > v_3 > \dots > v_{2020}$$
(by $\infty$, $y$ and $x$ in these inequalities we mean each number written on the cell with the corresponding label), and the numbers on empty cells are less than each of the numbers on marked cells.