2. The largest positive integer $n$ for which the inequality $\frac{9}{17}<\frac{n}{n+k}<\frac{8}{15}$ holds for a unique integer $k$ is $\qquad$
2. 144 .

From the problem, we know that $\frac{7}{8}<\frac{k}{n}<\frac{8}{9}$.
By the uniqueness of $k$, we have
$$
\begin{array}{l}
\frac{k-1}{n} \leqslant \frac{7}{8}, \text { and } \frac{k+1}{n} \geqslant \frac{8}{9} . \\
\text { Therefore, } \frac{2}{n}=\frac{k+1}{n}-\frac{k-1}{n} \geqslant \frac{8}{9}-\frac{7}{8}=\frac{1}{72} \\
\Rightarrow n \leqslant 144 .
\end{array}
$$

When $n=144$, from $\frac{7}{8}<\frac{k}{n}<\frac{8}{9}$, we get
$$
126<k<128 \text {. }
$$

Thus, $k$ can take the unique integer value 127. Therefore, the maximum value of the positive integer $n$ that satisfies the condition is 144.