3. It is known that the decimal part of the fraction $\frac{p}{q}$ contains the sequence of digits 171. Arrange all denominators $q$ that satisfy this condition in ascending order, then the sum of the smallest two denominators is $\qquad$ (here, decimals with a repeating cycle of 9 are written as having a repeating cycle of 0, for example: $0 . \dot{9}=1 . \dot{0}$)
3. 99 .

First, reduce the fraction $\frac{p}{q}$ so that the numerator and denominator are coprime, then multiply by a power of 10 to move the decimal point just before 171, and then change the integer part to 0. Throughout this process, the denominator of the fraction corresponding to the decimal either remains unchanged (no reduction or $q$ is coprime with 10) or decreases (reduction or $q$ is not coprime with 10). At this point, the smallest two values of the denominator of the new fraction corresponding to the decimal are the two smallest values of $q$ sought in this problem. Thus, without loss of generality,
$0.171 \leqslant \frac{p}{q}93.75$.
In summary, the two smallest values of $q$ are $35$ and $64$, and their sum is 99.