Intermediate fraction

Let $A = \frac {52}{303}, B = \frac {16}{91}$ ,

Then when $k > 0$ ,

$A < \frac {52 + 16k}{303 + 91k} = \frac {4(13+4k)}{303+91k} < B$

If we can find a $k$ , which let $13 + 4k$ be a factor of $303 + 91k$ ., then we get a fraction between $A$ and $B$ which has small sum of numerator and denominator.

Let $303 + 91k = (13 + 4k)n = 13n + 4kn \\ 303 - 13n = (4n - 91)k \\ k = \frac {303 - 13n}{4n - 91}$

Obviously there is positive integer solution $n = 23, k = 4$ , and:

$\frac {52 + 16k}{303 + 91k} = \frac{52 + 16 \times 4}{303 + 91 \times 4} = \frac {116}{667} = \frac {4 \times 29}{23 \times 29} = \frac {4}{23}$

I solved this one through trial and error, testing numerators and seeing what denominators would get them closest to the required range, until I found a fraction that worked.

$\frac{1}{6}<\frac{52}{303}<\frac{16}{91}<\frac{1}{5}$

$\frac{2}{12}<\frac{52}{303}<\frac{16}{91}<\frac{2}{11}$

$\frac{3}{18}<\frac{52}{303}<\frac{16}{91}<\frac{3}{17}$

$\frac{52}{303}<\frac{4}{23}<\frac{16}{91}$

$\frac{4}{23}$ works, and if we increase the numerator to $5$ , the denominator would increase as well, so the answer with the smallest denominator is indeed $\frac{4}{23}$ . Therefore the answer is $\boxed{4+23=27}$ .