Prove that all roots of $\sum_{r=1}^{70} \frac{1}{x-r} =\frac{5}{4} $ are real 
Prove that all roots of $$\displaystyle \sum_{r=1}^{70} \dfrac{1}{x-r} =\dfrac{5}{4} $$
are real

I encountered this question in my weekly test. I tried setting $\displaystyle \sum_{r=1}^{70} \dfrac{1}{x-r} = \dfrac{P'(x)}{P(x)} $ ; where $\displaystyle P(x) = \prod_{r=1}^{70} (x-r)$ and tried to use some inequality, but to no avail.
Can we also generalize this?

Find the condition such that
$$\displaystyle \sum_{r=1}^{n} \dfrac{1}{x-r} = a $$
has all real roots.
$n\in \mathbb{Z^+} \ ;  \ a\in \mathbb{R}$


Alternatively: suppose that the function has a complex root; call it $x+iy$; $y\neq 0$.
Then $$\sum_{r=1}^{70}\frac{1}{x-r+iy}=\frac{5}{4}$$
multiply by the conjugate to get
$$\sum_{r=1}^{70}\frac{x-r-iy}{(x-r)^2+y^2}=\frac{5}{4}$$
which implies
$$\operatorname{Im}\left( \sum_{r=1}^{70}\frac{x-r-iy}{(x-r)^2+y^2} \right)=0$$
or
$$\sum_{r=1}^{70}\frac{-y}{(x-r)^2+y^2}=0$$
which is a contradiction since each term has the same sign and is nonzero.
