Answer: 103

Solution:

If $ab + 2016 \ge 0,$ the first term is real, so the number is real exactly when $\sqrt{|a + b|} = 0,$ that is $b = -a.$ Then $ab + 2016 = 2016 - a^2 \ge 0$ forces $|a| \le 44,$ and the denominator $ab + 100 = 100 - a^2$ rules out $a = \pm 10.$ That gives $89 - 2 = 87$ pairs.

If $ab + 2016 \lt 0,$ then $\sqrt{ab + 2016} = i\sqrt{-ab - 2016},$ so the whole number is $\frac{\sqrt{-ab - 2016} - \sqrt{|a + b|}}{ab + 100}\,i,$ which is real exactly when $-ab - 2016 = |a + b|.$ Note $a + b = 0$ is impossible here since $a^2 = 2016$ has no integer solution. For $a + b \gt 0$ the equation becomes $ab + a + b + 2016 = 0,$ that is $(a + 1)(b + 1) = -2015,$ and for $a + b \lt 0$ it becomes $(a - 1)(b - 1) = -2015.$

Since $2015 = 5 \cdot 13 \cdot 31$ has $8$ positive divisors, $(a+1)(b+1) = -2015$ has $16$ ordered integer solutions, and $a + b \gt 0$ holds exactly when the positive factor is the larger in absolute value: $8$ solutions. Symmetrically the other case gives $8$ more. In all of these $ab + 100 = -1916 - |a + b| \ne 0.$ The total is $87 + 8 + 8 = 103.$