Solution:

Suppose that a certain number $N$ is special and has 4 digits. Then there will exist an odd number $d$ such that
$$N = (d-2)^2 + d^2 + (d+2)^2 = 3d^2 + 8$$
Since $d$ is odd, $N$ is also odd and therefore can only be $1111$, $3333$, $5555$, $7777$, $9999$. Now, however, $N$ cannot be $3333$ or $9999$, since these numbers are divisible by $3$ while $3d^2 + 8$ is not; $N$ also cannot be $1111$ or $7777$, since subtracting $8$ from these numbers gives a quantity (which should be $3d^2$) not divisible by $3$. The only possibility is therefore that $N = 5555$, from which we easily obtain $d = 43$. Hence $5555 = 41^2 + 43^2 + 45^2$ is the only special number with 4 digits.

Let us now come to the case of 2000 digits. The possibilities for the last digit of $d$ are $1, 3, 5, 7, 9$; for the last digit of $d^2$ they are $1, 5, 9$; for the last digit of $3d^2$ they are $3, 5, 7$; finally, for the last digit of $N = 3d^2 + 8$ they are $1, 3, 5$.

The case in which $N$ consists entirely of $3$'s is discarded since $N$ would be divisible by $3$ while $3d^2 + 8$ is not. The case in which $N$ is written with all $5$'s is discarded because $N-8$ would be written as $1998$ digits $5$ followed by $47$, and hence would not be divisible by $3$ (since the sum of its digits is $1998 \cdot 5 + 4 + 7$, which is not a multiple of $3$).

To examine the remaining case (all digits $1$) we set $d = 2k + 1$ in (1) and obtain $12k(k+1) = N - 11$. Since either $k$ or $k+1$ is an even number, the term on the left in this last equality is divisible by $8$, while the term on the right is
$$\underbrace{11 \ldots 11}_{1998 \text{ times}} 00 = 100 \cdot (\text{odd number})$$
and hence is not divisible by $8$.

This allows us to conclude that there do not exist special numbers with 2000 digits.