Solution:

To form $1$ cent we need $a = 1.$ Suppose the collection has $x$ ones, $y$ stamps of $b,$ and $z$ of $c.$ The value $b - 1$ must be made from ones alone, so $x \ge b - 1;$ the value $c - 1$ must be made from ones and $b$'s, so $x + yb \ge c - 1;$ and the total $x + yb + zc$ must be at least $1000.$ Conversely these three conditions suffice: with $x \ge b - 1$ the ones and $b$'s make every value up to $x + yb,$ and then $c$'s extend this to every value up to the total. So the optimum takes $x = b - 1,$ then the least $y$ with $x + yb \ge c - 1,$ then the least $z$ reaching $1000.$

For fixed $c,$ the count $f(1, b, c)$ is maximized at $b = c - 1$ (many ones, which cover value least efficiently), where the optimal collection is $c - 2$ ones, one stamp of $c - 1,$ and $\left\lceil \frac{1003 - 2c}{c} \right\rceil$ stamps of $c,$ totaling $c - 3 + \left\lceil \frac{1003}{c} \right\rceil.$ For $12 \le c \le 87$ this maximum is at most $96$ (it is $93$ at $c = 12,$ decreases in the middle, and returns to $96$ at $c = 87$), so no $b$ gives $97;$ a quick check of $c \le 10$ shows the possible counts skip $97$ there as well.

For $c = 11,$ taking $b = 7$ gives $6$ ones, one $7$ (reaching $13 \ge 10$), and $\left\lceil \frac{987}{11} \right\rceil = 90$ elevens: $f(1, 7, 11) = 6 + 1 + 90 = 97.$ For $c = 88$ and $c = 89,$ taking $b = 87$ gives $86$ ones, one $87$ (reaching $173$), and $\left\lceil \frac{827}{88} \right\rceil = \left\lceil \frac{827}{89} \right\rceil = 10$ stamps of $c,$ for $86 + 1 + 10 = 97$ in both cases. So the three least values of $c$ are $11, 88, 89,$ with sum $188.$