jmc

First, we can evaluate the sum across the first row, which gives $(n+1)+1+(n-1)=2n+1$. Evaluate the sum of the entries across the second row, $3+(2n-9)+n=3n-6$. Now, since we have a magic square, these two sums are equal. So $2n+1=3n-6$. Isolating $n$, we obtain $n=\boxed{7}$.

The square will look like: