We can start by multiplying both sides of the congruence by $ 10 $ and evaluating both sides modulo $ 997 $: \begin{align*} 10\cdot 100x &\equiv 10\cdot 1 \pmod{997} \\ 1000x &\equiv 10 \pmod{997} \\ 3x &\equiv 10 \pmod{997} \end{align*} Why multiply by $ 10 $? Well, as the computations above show, the result is to produce a congruence equivalent to the original congruence, but with a much smaller coefficient for $ x $. From here, we could repeat the same strategy a couple more times; for example, multiplying both sides by $ 333 $ would give $ 999x\equiv 2x $ on the left side, reducing the coefficient of $ x $ further. One more such step would reduce the coefficient of $ x $ to $ 1 $, giving us the solution. However, there is an alternative way of solving $ 3x\equiv 10\pmod{997} $. We note that we can rewrite this congruence as $ 3x\equiv -987\pmod{997} $ (since $ 10\equiv -987\pmod{997} $). Then $ -987 $ is a multiple of $ 3 $: specifically, $ -987 = 3\cdot (-329) $, so multiplying both sides by $ 3^{-1} $ gives $$x \equiv -329\pmod{997}.$$ This is the solution set to the original congruence. The unique three-digit positive solution is $$x = -329 + 997 = \boxed{668}.$$