The number of distinct positive divisors of an integer $N = \prod p_i^{e_i}$ with distinct primes $p_i$ is equal to $\prod (e_i + 1)$. Because $8$ has only the factorizations $8$, $4 \cdot 2$ and $2 \cdot 2 \cdot 2$, we have to consider three cases:

Case A: $N = p^7$
Case B: $N = p^3q$
Case C: $N = pqr$,
where $p, q, r$ are distinct primes. and these reduce to sub-cases.

In Case A we have $d_k = p^{k-1}$ and $2d_2d_5 = d_3d_4 + 3$ is equivalent to $p^5 = 3$, which does not hold for any $p$.

In Case B we consider the following four sub-cases
1) $q < p$
2) $p < q < p^2$
3) $p^2 < q < p^3$
4) $p^3 < q$.
These inequalities lead to identification of $d_2, d_3, d_4, d_5$ in each case as follows:
1) $d_2 = q$, $d_3 = p$, $d_4 = pq$, $d_5 = p^2$
2) $d_2 = p$, $d_3 = q$, $d_4 = p^2$, $d_5 = pq$
3) $d_2 = p$, $d_3 = p^2$, $d_4 = q$, $d_5 = p^3$
4) $d_2 = p$, $d_3 = p^2$, $d_4 = p^3$, $d_5 = q$
These give rise to one solution from sub-case 4) with $p = 3$ and $q = 41$ leading to $N = 3^3 \cdot 41 = 1107$. The first three cases would require $3$ to be divisible by $p^2$, which is not possible.

In Case C, w.l.o.g., we can assume $p < q < r$. We consider two sub-cases:
1) $r < pq$ and 2) $pq < r$.

1) $d_2 = p$, $d_3 = q$, $d_4 = r$, $d_5 = pq$

$2p^2q = qr + 3$, i.e. $q(2p^2 - r) = 3$ and so $q = 3$ and $2p^2 = r + 1$. As $p < q$ we get $p = 2$ and so $r = 7$. However, we now have $r = 7 > 6 = pq$ and no solutions arise.

2) $d_2 = p$, $d_3 = q$, $d_4 = pq$, $d_5 = r$

$2pr = pq^2 + 3$, i.e. $p(2r - q^2) = 3$. This gives $p = 3$. We have a solution if $r = (q^2 + 1)/2$. Using $q = 5, 7, 11$ gives $r = 13, 25, 61$, respectively. As $13 < pq = 15$ and $25$ is not prime, the smallest value of $q$ to yield a solution is $q = 11$. This gives $N = 2013$.

Larger values of $q$ for which $(q^2 + 1)/2$ is a prime (greater than $61$) yield other values of $N$ which will be bigger than $1107$ and $2013$. Thus these are the two smallest solutions.