Solution:

a) A direct check shows that the condition is satisfied when $a + b + c = d$. Let us assume that $a + b + c > d$. Then it is easy to see that
$$ab + cd > (d - a)(d - b)$$
We have analogously $bc + ad > (d - b)(d - c)$ and $ac + bd > (d - a)(d - c)$. Now the multiplication of these three inequalities gives a contradiction. Analogous arguments lead to a contradiction when $a + b + c < d$ and therefore $d = a + b + c$.

b) For a fixed $d$, $3 \leq d \leq n$, the equation $d = a + b + c$ has
$$\binom{d-1}{2} = \frac{(d-1)(d-2)}{2}$$
solutions. (This can be proved as follows. Write consecutively $d$ 1's. Then the number of the solutions is equal to the number of the ways one can put two separating lines in that sequence; for example $111|11 \ldots 11|1$ corresponds to $a = 3$, $b = d - 4$, $c = 1$.) This formula is true also for $d = 1$ and $d = 2$ since the equation has no solutions in these cases.

It remains to calculate
$$$$
$$\begin{aligned} \sum_{d=1}^{n} \frac{(d-1)(d-2)}{2} & = \frac{1}{2} \sum_{d=1}^{n} d^2 - \frac{3}{2} \sum_{d=1}^{n} d + n \\ & = \frac{1}{2} \cdot \frac{n(n+1)(2n+1)}{6} - \frac{3}{2} \cdot \frac{n(n+1)}{2} + n \\ & = \frac{n(n-1)(n-2)}{6} \end{aligned}$$