a) Considering the numbers from 1 to 20 we see that exactly three of them have 6 divisors: 12 (the divisors are 1, 2, 3, 4, 6, 12), 18 (1, 2, 3, 6, 9, 18), and 20 (1, 2, 4, 5, 10, 20). For 12 the sum $d_4 + d_5 = 4 + 6$ is divisible by the sum $d_2 + d_3 = 2 + 3$ and similarly, for 18 the sum $d_4 + d_5 = 6 + 9$ is divisible by $d_2 + d_3 = 2 + 3$. However, for 20 the sum $d_4 + d_5 = 5 + 10$ is not divisible by the sum $d_2 + d_3 = 2 + 4$. Thus, the smallest non-good number with exactly 6 factors is 20.

b) Take $N = 4p$, where $p$ is an arbitrary prime number larger than 4. Then $N$ has exactly 6 different divisors: 1, 2, 4, $p$, $2p$, $4p$, in increasing order. Indeed, as $N = 2^2p^1$, where 2 and $p$ are two different prime numbers, all of its divisors can be expressed as $2^i p^j$ where $i \le 2$ and $j \le 1$. From here we obtain exactly $3 \cdot 2 = 6$ choices: $i$ can be either 0, 1 or 2 and for every $i$ we have two choices for $j$: 0 or 1.
Here, $d_4 + d_5 = p + 2p = 3p$ is odd because $p > 2$ and thus is not divisible by an even number $d_2 + d_3 = 2 + 4 = 6$. Therefore, none of the numbers expressed as $N = 4p$ where $p > 4$ is a prime is good. As there are infinitely many prime numbers, there must also be infinitely many such numbers $N$.

Let us find all $N > 1$ which have exactly 6 divisors.

1) If $N = p^k$, where $p$ is a prime, then it has the divisors 1, $p$, ..., $p^k$, i.e. $k + 1$ divisors overall. Thus, all $N = p^5$ satisfy this requirement.

2) Let $N$ have two different prime divisors, i.e. $N = p^k q^l$. For $k \ge 2$ and $l \ge 2$ we see that $N$ has at least 9 different divisors: 1, $p$, $p^2$, $q$, $q^2$, $pq$, $p^2q$, $pq^2$, and $p^2q^2$. For $l = 1$, $N$ has the divisors 1, $p$, ..., $p^k$, and $q$, $pq$, ..., $p^kq$, i.e. $2(k + 1)$ divisors in total. Thus, all $N = p^2q$ satisfy the requirement.

3) Let $N$ have at least three prime divisors $p$, $q$, $r$. Then $N$ has at least 8 different divisors: 1, $p$, $q$, $r$, $pq$, $pr$, $qr$, and $pqr$, and we get no more numbers.

Let us now consider $N = p^5$ and $N = p^2q$ in more detail.

i) If $N = p^5$, then $d_i = p^{i-1}$ and $d_4 + d_5 = p^3 + p^4 = p^3(1 + p)$ is divisible by $d_2 + d_3 = p + p^2 = p(1 + p)$. Thus they are all good.

ii) If $N = p^2q$, where $q < p$, then $N$ has the divisors $1, q, p, pq, p^2$, and $p^2q$, in increasing order, and $d_4 + d_5 = pq + p^2 = p(q + p)$ is divisible by $d_2 + d_3 = q + p$ and thus, they are all good, too.

iii) If $N = p^2q$, where $p < q < p^2$, then $N$ has the divisors $1, p, q, p^2, pq$, and $p^2q$, in increasing order, and $d_4 + d_5 = p^2 + pq = p(p + q)$ is divisible by $d_2 + d_3 = p + q$. Thus, they are all good, too.

iv) Finally, let $N = p^2q$, where $q > p^2$. Then $N$ has the divisors $1, p, p^2, q, pq$, and $p^2q$, in increasing order and $d_4 + d_5 = q + pq = q(1 + p)$ is not divisible by $d_2 + d_3 = p + p^2 = p(1 + p)$ because the prime number $q$ cannot be divisible by another prime number $p$. Thus all these numbers have exactly 6 different divisors and they all are non-good. To get the smallest of these numbers, we have to take $p$ and $q$ as small as possible, i.e. $p = 2$ and $q = 5$ (to achieve $q > p^2 = 4$). Then $N = 2^2 \cdot 5 = 20$. Finally, there are infinitely many of these numbers $N$ because we have infinitely many choices for prime numbers $p$ and $q$ such that $q > p^2$. For example, we can take $p = 2$ and $q$ an arbitrary prime number bigger than 5. As there are infinitely many prime numbers, we have proven the statement.