Find all the pairs of integers $ (m, n)$  such that $ \sqrt {n +\sqrt {2016}} +\sqrt {m-\sqrt {2016}} \in \mathbb {Q}.$
1. We start with the given equation:
   \[
   \sqrt{n + \sqrt{2016}} + \sqrt{m - \sqrt{2016}} \in \mathbb{Q}
   \]
   This implies that the sum of the two square roots is a rational number.

2. Assume \( m \neq n \). Then there exists \( a \in \mathbb{Q} \) such that:
   \[
   \sqrt{n + \sqrt{2016}} + \sqrt{m - \sqrt{2016}} = a
   \]
   Squaring both sides, we get:
   \[
   n + \sqrt{2016} + m - \sqrt{2016} + 2\sqrt{(n + \sqrt{2016})(m - \sqrt{2016})} = a^2
   \]
   Simplifying, we have:
   \[
   n + m + 2\sqrt{(n + \sqrt{2016})(m - \sqrt{2016})} = a^2
   \]
   Isolating the square root term:
   \[
   2\sqrt{(n + \sqrt{2016})(m - \sqrt{2016})} = a^2 - n - m
   \]
   Squaring both sides again:
   \[
   4(n + \sqrt{2016})(m - \sqrt{2016}) = (a^2 - n - m)^2
   \]
   Expanding the left-hand side:
   \[
   4(nm - n\sqrt{2016} + m\sqrt{2016} - 2016) = (a^2 - n - m)^2
   \]
   Since \( \sqrt{2016} \) is irrational, the left-hand side contains irrational terms, while the right-hand side is rational. This leads to a contradiction unless the irrational terms cancel out, which implies \( m = n \).

3. Therefore, we must have \( m = n \). The equation simplifies to:
   \[
   2\sqrt{n^2 - 2016} = a^2 - 2n
   \]
   Let \( k = \sqrt{n^2 - 2016} \). Then:
   \[
   2k = a^2 - 2n
   \]
   Solving for \( k \):
   \[
   k = \frac{a^2 - 2n}{2}
   \]
   Since \( k \) must be an integer (as \( n^2 - 2016 \) is an integer), \( a^2 - 2n \) must be even.

4. We have:
   \[
   n^2 - 2016 = k^2
   \]
   This can be factored as:
   \[
   (n - k)(n + k) = 2016
   \]
   We need to find pairs of integers \( (n, k) \) such that the product is 2016. The factor pairs of 2016 are:
   \[
   (1, 2016), (2, 1008), (3, 672), (4, 504), (6, 336), (7, 288), (8, 252), (9, 224), (12, 168), (14, 144), (16, 126), (18, 112), (21, 96), (24, 84), (28, 72), (32, 63), (36, 56), (42, 48)
   \]
   Solving for \( n \) in each case:
   \[
   n = \frac{(n - k) + (n + k)}{2} = \frac{2016 + 1}{2}, \frac{1008 + 2}{2}, \ldots
   \]
   This gives us the possible values of \( n \):
   \[
   n = 1008.5, 505, 337.5, 254, 171, 130, 90, 79, 71, 65, 54, 50, 46, 45
   \]
   Only integer values are valid, so:
   \[
   n = 505, 254, 130, 65, 50, 46, 45
   \]

5. We need to check which of these values satisfy the initial inequality \( a^2 \ge 2n \):
   - For \( n = 505 \), \( a^2 \ge 1010 \) (possible)
   - For \( n = 254 \), \( a^2 \ge 508 \) (possible)
   - For \( n = 130 \), \( a^2 \ge 260 \) (possible)
   - For \( n = 65 \), \( a^2 \ge 130 \) (possible)
   - For \( n = 50 \), \( a^2 \ge 100 \) (possible)
   - For \( n = 46 \), \( a^2 \ge 92 \) (possible)
   - For \( n = 45 \), \( a^2 \ge 90 \) (possible)

The final answer is \( \boxed{ (m, n) = (505, 505), (254, 254), (130, 130), (65, 65), (50, 50), (46, 46), (45, 45) } \)