How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of is a perfect square?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$
Let $N = 10t + u$, where $t$ is the tens digit and $u$ is the units digit.
The condition of the problem is that $10t + u + 10u + t$ is a perfect square.
Simplifying and factoring, we want $11(t+u)$ to be a perfect square.
Thus, $t+u$ must at least be a multiple of $11$, and since $t$ and $u$ are digits, the only multiple of $11$ that works is $11$ itself.
Thus, $(t,u) = (2,9)$ is the first solution, and $(t,u) = (9,2)$ is the last solution.  There are $8$ solutions in total, leading to answer $\boxed{E}$.