Solution:

We shall say that the order of Spas is $1$, the order of his parents is $2$, the order of their parents is $3$, etc. Write the number of the heads of any dragon in ternary base: Spas has $1$ head, his mother $10$ heads, his father $11$ heads, etc. It follows by induction on the order of the dragons that the heads of the dragons of order $n$ are $n$-digit numbers whose ternary representation contains no digit $2$.

It follows that if two dragons have $a + b = k$ ($a > b$) heads in total and $k$ is written in ternary base, then $a$ and $b$ have $0$ at the positions where $k$ has $0$ and $1$ at the positions where $k$ has $2$. Moreover, if $k$ has $1$ at some position, then one of $a$ and $b$ has $1$ and the other one has $0$ at the same position.

So if the ternary representation of $k$ contains no $1$, then $a = b$ and hence $k$ is not a good number. If this representation contains at least two $1$'s ($k = \ldots 1 \ldots 1 \ldots$), we have two possibilities: $(a = \ldots 1 \ldots 1 \ldots, b = \ldots 0 \ldots 0 \ldots)$ and $(a = \ldots 1 \ldots 0 \ldots, b = \ldots 0 \ldots 1 \ldots)$. Such a $k$ is not a good number except for the case when $k$ contains exactly two $1$'s and the other digits are $0$.

If $k$ contains exactly one $1$, then $a$ has $1$ and $b$ has $0$ at this position and the other digits of $a$ and $b$ are uniquely determined. The only exception is the case when the remaining digits of $k$ are $0$. All the numbers with one $1$ and at least one $2$ are good.

Since $2003 = 2202012_{3}$, the number $2003$ is good.

Let us count all good numbers with at most $7$ digits. One has $\binom{7}{2} = 21$ numbers with exactly two $1$'s and $0$ at the other positions. There are $2^{6} - 1 = 63$ non-zero numbers with at most $6$ digits equal to $0$ or $2$. There exist $7$ possibilities to put $1$ in such numbers. So we get $7 \cdot 63 = 441$ numbers and adding the above $21$ numbers, we obtain $462$ good numbers with at most $7$ digits.

We shall count the $7$-digit good numbers greater than $2202012_{3}$. They are $2202021_{3}, 2202100_{3}, 2202102_{3}, 2202120_{3}, 2202122_{3}, 2202201_{3}, 2202210_{3}$, $2202212_{3}, 2202221_{3}$, $16$ numbers of the form $\overline{221mnpq}_{3}$, where $m, n, p, q \in \{0 ; 2\}$ and $4 \cdot 8 = 32$ numbers of the form $\overline{222mnpq}_{3}$, where exactly one of the digits $m, n, p, q$ is equal to $1$. Thus, there are $1 + 9 + 16 + 32 = 58$ $7$-digit good numbers greater than $2202012_{3}$.

Hence we have $462 - 58 = 404$ good numbers smaller than $2003$.