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.
$n\in \mathbb{Z^+} \ ;  \ a\in \mathbb{R}$


Hint: Note that the derivative of the left-hand side, when it exists, is negative. So our function is decreasing in any interval $(k,k+1)$ where $1\le k\le 70-1$. 
In this interval, our function is very large positive when $x$ is a tiny bit larger than $k$, and very large negative when $x$ is a tiny bit smaller than $k+1$. So by the Intermediate Value Theorem our equation has a root between $k$ and $k+1$.
