Suppose that a number $q$ has the property that there are exactly 19 integers $n$ with $\sqrt{q}<n<q$. Suppose that these 19 integers are $m, m+1, m+2, \ldots, m+17, m+18$. Then $\sqrt{q}<m<m+1<m+2<\cdots<m+17<m+18<q$. This tells us that $q-\sqrt{q}>(m+18)-m=18$ because $q-\sqrt{q}$ is as small as possible when $q$ is as small as possible and $\sqrt{q}$ is as large as possible. Also, since this is exactly the list of integers that is included strictly between $\sqrt{q}$ and $q$, then we must have $m-1 \leq \sqrt{q}<m<m+1<m+2<\cdots<m+17<m+18<q \leq m+19$. In other words, neither $m-1$ nor $m+19$ can satisfy $\sqrt{q}<n<q$. This tell us that $q-\sqrt{q} \leq(m+19)-(m-1)=20$. Therefore, we have that $18<q-\sqrt{q} \leq 20$. Next, we use $18<q-\sqrt{q} \leq 20$ to get a restriction on $q$ itself. To have $q-\sqrt{q}>18$, we certainly need $q>18$. But if $q>18$, then $\sqrt{q}>\sqrt{18}>4$. Furthermore, $q-\sqrt{q}>18$ and $\sqrt{q}>4$ give $q-4>q-\sqrt{q}>18$ and so $q>22$. Next, note that $q-\sqrt{q}=\sqrt{q}(\sqrt{q}-1)$. When $q$ is larger than 1 and increases, each factor $\sqrt{q}$ and $\sqrt{q}-1$ increases, so the product $q-\sqrt{q}$ increases. When $q=25, q-\sqrt{q}=25-5=20$. Since we need $q-\sqrt{q} \leq 20$ and since $q-\sqrt{q}=20$ when $q=25$ and since $q-\sqrt{q}$ is increasing, then for $q-\sqrt{q} \leq 20$, we must have $q \leq 25$. Therefore, $18<q-\sqrt{q} \leq 20$ tells us that $22<q \leq 25$. So we limit our search for $q$ to this range. When $q=22, \sqrt{q} \approx 4.69$, and so the integers $n$ that satisfy $\sqrt{q}<n<q$ are $5,6,7, \ldots, 20,21$, of which there are 17. When $22<q \leq 23$, we have $4<\sqrt{q}<5$ and $22<q \leq 23$, which means that the integers $n$ that satisfy $\sqrt{q}<n<q$ are $5,6,7, \ldots, 20,21,22$, of which there are 18. When $23<q \leq 24$, we have $4<\sqrt{q}<5$ and $23<q \leq 24$, which means that the integers $n$ that satisfy $\sqrt{q}<n<q$ are $5,6,7, \ldots, 20,21,22,23$, of which there are 19. When $24<q<25$, we have $4<\sqrt{q}<5$ and $24<q<25$, which means that the integers $n$ that satisfy $\sqrt{q}<n<q$ are $5,6,7, \ldots, 20,21,22,23,24$, of which there are 20. When $q=25, \sqrt{q}=5$ and so the integers that satisfy $\sqrt{q}<n<q$ are $6,7, \ldots, 20,21,22,23,24$, of which there are 19. Therefore, the numbers $q$ for which there are exactly 19 integers $n$ that satisfy $\sqrt{q}<n<q$ are $q=25$ and those $q$ that satisfy $23<q \leq 24$. Finally, we must determine the sum of all such $q$ that are of the form $q=\frac{a}{b}$ where $a$ and $b$ are positive integers with $b \leq 10$. The integers $q=24$ and $q=25$ are of this form with $a=24$ and $a=25$, respectively, and $b=1$. The $q$ between 23 and 24 that are of this form with $b \leq 4$ are $23 \frac{1}{2}=\frac{47}{2}, 23 \frac{1}{3}=\frac{70}{3}, 23 \frac{2}{3}=\frac{71}{3}, 23 \frac{1}{4}=\frac{93}{4}, 23 \frac{3}{4}=\frac{95}{4}$. Notice that we don't include $23 \frac{2}{4}$ since this is the same as the number $23 \frac{1}{2}$. We continue by including those satisfying $5 \leq b \leq 10$ and not including equivalent numbers that have already been included with smaller denominators, we obtain $23 \frac{1}{2}, 23 \frac{1}{3}, 23 \frac{2}{3}, 23 \frac{1}{4}, 23 \frac{3}{4}, 23 \frac{1}{5}, 23 \frac{2}{5}, 23 \frac{3}{5}, 23 \frac{4}{5}, 23 \frac{1}{6}, 23 \frac{5}{6}, 23 \frac{1}{7}, 23 \frac{2}{7}, 23 \frac{3}{7}, 23 \frac{4}{7}, 23 \frac{5}{7}, 23 \frac{6}{7}, 23 \frac{1}{8}, 23 \frac{3}{8}, 23 \frac{5}{8}, 23 \frac{7}{8}, 23 \frac{1}{9}, 23 \frac{2}{9}, 23 \frac{4}{9}, 23 \frac{5}{9}, 23 \frac{7}{9}, 23 \frac{8}{9}, 23 \frac{1}{10}, 23 \frac{3}{10}, 23 \frac{7}{10}, 23 \frac{9}{10}$. There are 31 numbers in this list. Each of these 31 numbers equals 23 plus a fraction between 0 and 1. With the exception of the one number with denominator 2, each of the fractions can be paired with another fraction with the same denominator to obtain a sum of 1. Therefore, the sum of all of these $q$ between 23 and 24 is $31(23)+\frac{1}{2}+15(1)=728 \frac{1}{2}$, because there are 31 contributions of 23 plus the fraction $\frac{1}{2}$ plus 15 pairs of fractions with a sum of 1. Finally, the sum of all $q$ of the proper form for which there are exactly 19 integers that satisfy $\sqrt{q}<n<q$ is $728 \frac{1}{2}+25+24=777 \frac{1}{2}$.