Answer: 188

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 $c,$ no $b$ can require more stamps than $b=c-1.$ Indeed, for any $2\le b\lt c,$ take $b-1$ ones and $c-b$ stamps of $b.$ These $c-1$ lower-denomination stamps have total value $$b-1+b(c-b)\ge 2c-3,$$ because the difference is $(b-2)(c-b-1)\ge0.$ Adding $\left\lceil\frac{1003}{c}\right\rceil-2$ stamps of $c$ therefore gives a working collection of at most $c-3+\left\lceil\frac{1003}{c}\right\rceil$ stamps. Equality is attained when $b=c-1:$ the mandatory $c-2$ ones and one $c-1$ stamp have value $2c-3,$ and stamps of value at most $c$ cannot reach $1000$ with fewer than $$c-3+\left\lceil\frac{1003}{c}\right\rceil$$ stamps in total.

For $12\le c\le87,$ the endpoint bounds $c(99-c)\ge1003$ give $\left\lceil\frac{1003}{c}\right\rceil\le99-c,$ so this maximum is at most $96$ and no $b$ gives $97.$ For $c\le10,$ any $97$ stamps have total value at most $97c\le970,$ so they cannot cover every value through $1000.$

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 $188.$