To solve the inequality \( n + 7 \) is divisible by \( 2n + 9 \), we can denote the expression as:

\[
\frac{n + 7}{2n + 9} \text{ is an integer}
\]

This implies there exists some integer \( k \) such that:

\[
n + 7 = k(2n + 9)
\]

Rearranging this gives:

\[
n + 7 = 2kn + 9k
\]
\[
n - 2kn = 9k - 7
\]
\[
n(1 - 2k) = 9k - 7
\]
\[
n = \frac{9k - 7}{1 - 2k}
\]

For \( n \) to be an integer, \( 9k - 7 \) must be divisible by \( 1 - 2k \). You can test various integer values for \( k \) to find possible values for \( n \).

The conditions that \( k \) can take will determine the possible values of \( n \). Specifically, when \( 1 - 2k \) is a factor of \( 9k - 7 \) leads us to check cases for integer solutions.

Would you like to go through specific values for \( k \) to find corresponding \( n \)?