Given that there are only three positive integers between the fractions $\frac{112}{19}$ and $\frac{112+x}{19+x}$. Find the sum of all possible integer values of $x$.
Notice that $50$ or $x<-19$.
Therefore, the three positive integers between $\frac{112+x}{19+x}$ and $\frac{112}{19}$ are $3,4,5$.
From $2<\frac{112+x}{19+x}<3$, we get $\frac{55}{2}<x<74$.
Since $x$ is an integer greater than 0, thus, $x=28,29, \cdots, 73$.
Therefore, the sum of these numbers is
$$
\frac{(28+73) \times 46}{2}=2323 \text {. }
$$
(2) If $\frac{112}{19}<\frac{112+x}{19+x}$, then $-19<x<0$.

Therefore, the three positive integers between $\frac{112}{19}$ and $\frac{112+x}{19+x}$ are $6,7,8$.
From $8<\frac{112+x}{19+x}<9$, we get $-\frac{59}{8}<x<-\frac{40}{7}$.
Since $-19<x<0$, thus, $x=-7,-6$.
Therefore, the sum of these numbers is -13.
Hence, the sum of all possible values of the integer $x$ is 2310.