There are integers $m$ and $n$ so that $9 +\sqrt{11}$  is a root of the polynomial $x^2 + mx + n.$ Find $m + n.$
1. Given that \(9 + \sqrt{11}\) is a root of the polynomial \(x^2 + mx + n\), we can use the fact that if \(a + b\) is a root of a polynomial with rational coefficients, then its conjugate \(a - b\) must also be a root. Therefore, \(9 - \sqrt{11}\) is also a root of the polynomial.

2. The polynomial with roots \(9 + \sqrt{11}\) and \(9 - \sqrt{11}\) can be written as:
   \[
   (x - (9 + \sqrt{11}))(x - (9 - \sqrt{11}))
   \]

3. Expanding this product:
   \[
   (x - 9 - \sqrt{11})(x - 9 + \sqrt{11}) = (x - 9)^2 - (\sqrt{11})^2
   \]

4. Simplifying the expression:
   \[
   (x - 9)^2 - 11 = x^2 - 18x + 81 - 11 = x^2 - 18x + 70
   \]

5. Comparing this with the polynomial \(x^2 + mx + n\), we get:
   \[
   m = -18 \quad \text{and} \quad n = 70
   \]

6. Therefore, \(m + n = -18 + 70 = 52\).

The final answer is \(\boxed{52}\).