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

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.

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 $k+1$.