The answer is $2022^2 + 1 = 4088485$.
First we show that $2022^2$ is insufficient. Suppose there are $2022^2$ beetles in the top left $2022 \times 2022$ square grid. This square grid can be partitioned into many $2 \times 2$ square grids in the natural way. If the beetles in each $2 \times 2$ square grid move in the anticlockwise direction as shown, then no beetle will ever meet.

Next, suppose there are at least $2022^2 + 1$ beetles. We label all the unit squares as follows. In the odd rows, we label the unit squares $ABABAB\cdots ABA$. In the even rows, we label the unit squares $CDCDCD\cdots CDC$. Note that there are $1011^2$ unit squares with label $D$.
As there are at least $2022^2+1$ beetles in total, we can always find at least $1011^2+1$ beetles lying in unit squares with the same label by the pigeonhole principle. In each move, either all beetles move in the order $A \to B \to D \to C \to A$ or all beetles move in the order $A \to C \to D \to B \to A$. In any case, these $1011^2 + 1$ beetles must move to unit squares with label $D$ simultaneously in a certain move. As there are only $1011^2$ unit squares with label $D$, two of the beetles must meet at a certain unit square. This completes the proof.