1.28 Given $x, y \in N$, find the largest $y$ value such that there exists a unique $x$ value satisfying the following inequality:
$$
\frac{9}{17}<\frac{x}{x+y}<\frac{8}{15} \text {. }
$$
(Wuhan, Hubei Province, China Mathematical Summer Camp, 1987)
[Solution 1] Since $x, y \in N$, the given inequality can be transformed into
$$
\left\{\begin{array}{l}
8 x-9 y>0, \\
7 x-8 y<0 .
\end{array}\right.
$$

For the interval $\left(\frac{9 y}{8}, \frac{8 y}{7}\right)$ to contain a unique integer $x$, the necessary and sufficient conditions are:
(1) If $\frac{8 y}{7} \in N$, then $\frac{8 y}{7}=\left[\frac{9 y}{8}\right]+2$, or
(2) If $\frac{8 y}{7} \notin N$, then $\left[\frac{8 y}{7}\right]=\left[\frac{9 y}{8}\right]+1$.
Let $y=8 t+r, t \in N, r=0,1,2, \cdots, 7$.
From (1), we have
$$
9 t+\frac{t+8 r}{7}=9 t+\left[\frac{9 r}{8}\right]+2
$$

which simplifies to
$$
\frac{t+r}{7}=\left[\frac{r}{8}\right]+2
$$

Since $\left[\frac{r}{8}\right]=0$, we have
$$
\frac{t+r}{7} \in N, \quad t \in N .
$$

Thus,
$$
t+r=14.
$$

To maximize $y$, we take $r=0$.
Therefore,
$$
t=14, \quad y=112.
$$

The corresponding $x=127$ is the unique integer.
From (2), we have
$$
\left[\frac{t+r}{7}\right]=\left[\frac{r}{8}\right]+1=1,
$$

so we can take the maximum $t+r=13$,
To maximize $y$, we take $r=0, \quad t=13$,
In this case, $y$ is maximized at 104, with the corresponding $x=118$.
From the above, the maximum value of $y$ is 112.
[Solution 2] The original inequality can be transformed into
$$
\frac{15}{8}<\frac{x+y}{x}<\frac{17}{9}
$$

which simplifies to
$$
\begin{array}{c}
\frac{7}{8}<\frac{y}{x}<\frac{8}{9} \\
63 y<56 x<64 y .
\end{array}
$$

Thus, the problem is to find the largest open interval $(63 y, 64 y)$ that contains only one multiple of 56. Since the interval length is $y$, it contains $y-1$ integers.
If $y-1 \geqslant 2 \cdot 56$, then the interval must contain at least 2 multiples of 56.
Therefore, $y-1<2 \cdot 56$, which means $y \leqslant 112$.
When $y=112$, we have
$$
63 \cdot 112=126 \cdot 56<127 \cdot 56<128 \cdot 56=64 \cdot 112,
$$

where $x=127$, so $y=112$ is the solution.