Since $\lfloor y \rfloor$ and $2022$ are integers, the equation $2x + \lfloor y \rfloor = 2022$ implies that $2x$ is also an integer, so $\lfloor 2x \rfloor = 2x$. Thus we can eliminate the unknown $x$ by subtracting the first equation of the system from the second one. We get
$$3y - \lfloor y \rfloor = 1. \qquad (1)$$
Thanks to (1), $3y$ is an integer, so it has (according to its remainder after division by three) one of the forms $3k$, $3k + 1$ or $3k + 2$, where $k$ is an integer. From this it follows that either $y = k$, or $y = k + \frac{1}{3}$, or $y = k + \frac{2}{3}$, where $k = \lfloor y \rfloor$. We now discuss these three cases.

* In the case of $y = k$, (1) becomes $3k - k = 1$ with the non-integer solution $k = \frac{1}{2}$.

* In the case of $y = k + \frac{1}{3}$, (1) is the equation $(3k + 1) - k = 1$ with a solution $k = 0$, which corresponds to $y = \frac{1}{3}$. The original system of equations is then apparently fulfilled, precisely when $2x = 2022$, i.e. $x = 1011$.

* In the case of $y = k + \frac{2}{3}$, (1) is the equation $(3k + 2) - k = 1$ with a non-integer solution $k = -\frac{1}{2}$.

Conclusion. The only solution of the given system is the pair $(x, y) = (1011, \frac{1}{3})$.

The verbal definition attached to the problem formulation tells us that $\lfloor a \rfloor$ is an integer for which $\lfloor a \rfloor \le a$ and at the same time $\lfloor a \rfloor + 1 > a$. So, estimates $a - 1 < \lfloor a \rfloor \le a$ are valid for every real number $a$. According to them, we get from the first equation of the given system
$$2022 \le 2x + y < 2023. \qquad (2)$$
Similarly, from the second equation follows
$$2023 \le 3y + 2x < 2024. \qquad (3)$$
We can combine these inequalities in two ways. Combining the second part of (2) with the first part of (3) we obtain $2x + y < 2023 \le 3y + 2x$, whence from the comparison of the outermost expressions follows $y > 0$. If we modify the first part of (2) to $2024 \le 2x + y + 2$, then together with the second part of (3) we obtain $3y + 2x < 2024 \le 2x + y + 2$. This time $y < 1$ follows from the comparison of the outermost expressions.

Together, we got $0 < y < 1$, so $\lfloor y \rfloor = 0$ holds. Thanks to this, the first equation of the original system is reduced to the form $2x = 2022$, which is satisfied only for $x = 1011$. By inserting it into the second equation, we get $3y + 2022 = 2023$ with the only solution $y = \frac{1}{3}$ which indeed satisfies the condition $\lfloor y \rfloor = 0$ used in the first equation. The pair $(x, y) = (1011, \frac{1}{3})$ is therefore the only solution of the given system.