Let $N$ be the number on the blackboard. Then, the number written by Carlos in his notebook is $C = 4N + 30$. Call $D$ the number written by Dora.
Since $D = C \ge N$ is a four digit number, the same holds for $C$; then, the first digit of $N$ is not greater than $2$, since, otherwise, $C \ge 4 \cdot 3000 = 12000$. On the other hand, it is clear that $4N + 30$ is even. So, the last digit of $D$, which is the first digit of $N$, is even (and greater than $0$). Combining both previous remarks, we conclude that the first digit of $N$ is $2$. Since $D \ge 4N \ge 4 \cdot 2000 = 8000$, we also deduce that the first digit of $D$ is at least $8$.
As adding $30$ does not change the last digit of a number, we have that $D$ and $4N$ have the same last digit, which is $2$. In order for $4N$ to end with $2$, $N$ has to end with $2$ or $8$. But the last digit of $N$ is the first digit of $D$, which we know is at least $8$. Then, the only possibility is that $N$ ends with $8$.
Thus, the number on the blackboard is of the form $N = \overline{2ab8} = 2000 + 100a + 10b + 8$, whereas $D = \overline{8ba2} = 8000 + 100b + 10a + 2$. Now, the relation $D = C$ can be restated as
$$8000 + 100b + 10a + 2 = 8000 + 400a + 40b + 32 + 30,$$
which simplifies to $60b - 60 = 390a$, and dividing by $30$, we obtain
$$2b - 2 = 13a.$$
Since $b$ is a digit, we have $-2 \le 2b - 2 \le 2 \cdot 9 - 2 = 16$. The only multiples of $13$ in this interval are $0$ and $13$, but taking into account that $2b - 2$ is even, we conclude that the only possibility is that $2b - 2 = 0$. Then, $b = 1$ and $a = 0$.
Therefore, $N = 2018$, which satisfies the condition and is the unique solution.