a. Suppose there exist $a, b \in \mathbb{Z}^+$ such that $a+b = 2003$ and $2003 \mid ab$. Since $2003$ is a prime, we must have $2003 \mid a$ or $2003 \mid b$. WLOG assume $2003 \mid a$. Then we have $a+b \ge 2003+1 > 2003$, which is a contradiction. Therefore, $2003$ is not good.

b. No. Suppose there exist $a, b \in \mathbb{Z}^+$ such that $a+b = 2002$ and $2002 \mid ab$. Note that $2002 = 2 \times 7 \times 11 \times 13$. For each prime $p \mid 2002$, we have $p \mid ab$, and so $p \mid a$ or $p \mid b$. In either case, since $p \mid a+b$, we must have $p \mid a$ and $p \mid b$. As this holds for $p = 2, 7, 11, 13$, we must have $2002 \mid a, b$. This yields $a+b \ge 2002+2002 > 2002$, which is a contradiction. Therefore, $2002$ is not good.

c. All positive integers of the form $4p_1p_2\cdots p_s$ or $8p_1p_2\cdots p_s$ where $p_1, p_2, \dots, p_s$ are distinct odd primes (possibly $s=0$) are good but not very good.

In view of the proof of part (b), all squarefree positive integers are not good. For any non-squarefree positive integer $n$, we can write $n = p^2m$ for some prime $p$ and positive integer $m$. Since we can take $a = pm$ and $b = (p-1)pm$ such that
$$a + b = p^2m = n$$
and
$$ab = (p-1)p^2m^2 = (p-1)mn,$$
we know that $n$ is good by definition. Also, note that $a$ and $b$ are distinct unless $p=2$. This shows $n$ is very good if $n$ is divisible by the square of an odd prime.

Thus, it remains to consider $n = 2^k m$ for some $k \ge 2$ and odd $m$ which is squarefree. If $k \ge 4$, we can take $a = 2^{k-2}m$ and $b = 3 \cdot 2^{k-2}m$ such that
$$a + b = 2^k m = n$$
and
$$ab = 3 \cdot 2^{2k-4}m^2 = 3 \cdot 2^{k-4}mn.$$
Since $a \neq b$, this shows $n$ is very good.

Lastly, if $k=2,3$, we show that $n$ is not very good. Indeed, suppose $a+b=n$ and $n|ab$. As in part (b), we deduce $m|a$ and $m|b$. Let $a=mc$ and $b=md$. Then we need $c+d=2^k$ and $2^k|cd$. It is routine to check that there is no solution except $c=d$ when $k=2,3$ (note that there are only a few pairs of $(c,d)$ satisfying $c+d=2^k$). Therefore, $n$ is not very good.