Three, given $x, y \in N$, find the largest $y$ value such that there exists a unique $x$ value satisfying the following inequality:
$$
\frac{9}{17}<\frac{x}{x+y}<\frac{8}{15} \text {. }
$$
Solve the original system of inequalities $\Longleftrightarrow\left\{\begin{array}{l}8 x-9 y>0, \\ 7 x-8 y>0\end{array}\right.$ $\longleftrightarrow \frac{9 y}{8}<x < \frac{8 y}{7}$.
For the interval $\left(\frac{9 y}{\circ}, \frac{8 y}{7}\right)$ to contain a unique integer, the necessary and sufficient condition is either $1^{\circ}$ or $2^{\circ}$:
$1^{\circ} \quad$ If $\frac{8 y}{7} \in \mathbb{N}$, then $\frac{8 y}{7}=\left[\frac{9 y}{8}\right]+2$;
$2^{\circ}$ If $\frac{8 y}{7} \notin \mathbb{N}$, then $\left[\frac{8 y}{7}\right]=\left[\frac{9 y}{8}\right]+1$.
For convenience, let $y=8 t+r, t \in \mathbb{N}, r=0$, $1,2, \cdots, 7$.

From $1^{\circ}$, we have $9 t+\frac{t+8 r}{7}=9 t+\left[\frac{9 r}{8}\right]+2$, which simplifies to $\frac{t+r}{7}=\left[\frac{r}{8}\right]+2$.

Here, $\left[\frac{r}{8}\right]=0$, and $\frac{t+r}{7} \in \mathbb{N}$. In this case, $r=0$. When $t$ is 14, $y$ reaches its maximum value of 112. The corresponding $x=127$, which is unique.

From $2^{\circ}$, similarly, we have $\left[\frac{t+r}{7}\right]=\left[\frac{r}{8}\right]+1$ $=1$. Since $\left(\frac{t+r}{7}\right) \in \mathbb{N}$, we can take $t+r=13$, i.e., $t=13, r=0$, at which point $y_{\max }=104$. The corresponding $x=118$.
Combining the above, we find that the required $y_{\max }=112$.