Solution:

The answer is 3. In general, if at a certain point the grasshopper is at $(a, b)$, at the previous step it was at a point $(x, y)$ such that $(a, b) = (x+y, y)$ or $(a, b) = (x, x+y)$, that is, at $(a-b, b)$ or at $(a, b-a)$; moreover, if it started from a point $(x_0, y_0)$ with positive coordinates, as the statement assures us, then at every moment it is located at a point with positive coordinates, and therefore the position preceding $(a, b)$ is uniquely determined and is $(a-b, b)$ if $a > b$, $(a, b-a)$ if $b > a$.

It follows that the positions of the grasshopper preceding $(2021,2050)$, going backwards, are: $(2021,2050-2021) = (2021,29)$, $(2021-29,29)$, $(2021-2 \cdot 29,29)$, $\ldots$, $(2021-69 \cdot 29,29) = (20,29)$, $(20,29-20) = (20,9)$. The grasshopper could therefore have started from $(20,9)$; if it did not, it could have made one more jump and started from $(20-9,9) = (11,9)$ or two more jumps and started from $(11-9,9) = (2,9)$. If the grasshopper had made even more jumps, it would have started from a point whose second coordinate is at most 7, which does not satisfy the conditions of the statement. There are therefore 3 possibilities for $(x_0, y_0)$.