let wlog order the a's as wlog $ \ a_{1}<a_{2}<...a_{n}$ let among them $ \ a_{1}<a_{2}<...a_{m}<0<a_{m+1}..<a_{n}$ notice 0 is a root then see if we approach a_{1} from right it blows to +infinity but as we approach a_{2} from left it blows to -infinity. but in $ \ (a_{1},a_{2})$ the fn is continuous so atleast at one point in the interval it take value n. apply same logic upto a_{m}. between a_{m} and a_{m+1} automatically a root $ \ 0 $. then apply again from a_{m+1}to thye next a etc.... so we get atleast $ \ n-1 $ real roots done

amparvardi wrote: If $a_i \ (i = 1, 2, \ldots, n)$ are distinct non-zero real numbers, prove that the equation \[\frac{a_1}{a_1-x} + \frac{a_2}{a_2-x}+\cdots+\frac{a_n}{a_n-x} = n\] has at least $n - 1$ real roots. Wlog say $a_1<a_2<...<a_n$ $f(x)=\frac{a_1}{a_1-x} + \frac{a_2}{a_2-x}+\cdots+\frac{a_n}{a_n-x}$ is continuous over $(a_k,a_{k+1})$ $\forall k\in[1,n-1]$ If $a_k<0$ and $a_{k+1}<0$ : $\lim_{x\to a_k^+}f(x)=+\infty$ and $\lim_{x\to a_{k+1}^-}f(x)=-\infty$ and at least one root in $(a_k,a_{k+1)}$ If $a_k<0$ and $a_{k+1}>0$ : $f(0)=n$ and at least one root in $(a_k,a_{k+1)}$ If $a_k>0$ and $a_{k+1}>0$ : $\lim_{x\to a_k^+}f(x)=-\infty$ and $\lim_{x\to a_{k+1}^-}f(x)=+\infty$ and at least one root in $(a_k,a_{k+1)}$ So at least one root of $f(x)=n$ in each interval $(a_k,a_{k+1})$ for any positive integer $k\in[1,n-1]$ Hence the result.

If all $a_j$'s had been of same sign the problem would have been very easy.

Write the equation \[\frac{a_1}{a_1-x} + \frac{a_2}{a_2-x}+\cdots+\frac{a_n}{a_n-x} = n\] as \[\frac{x}{a_1-x} + \frac{x}{a_2-x}+\cdots+\frac{x}{a_n-x} = 0.\] Consider the polynomial $p(x) = (x-a_1)(x-a_2)\cdots (x-a_n)$; then \[\frac{x}{a_1-x} + \frac{x}{a_2-x}+\cdots+\frac{x}{a_n-x} = -\dfrac {xp'(x)}{p(x)},\] so the problem comes to showing $xp'(x) = 0$ has at least $n-1$ real roots. But $p'$ indeed does have $n-1$ distinct real roots, by the definition of $p$. Moreover, if all $a_i$ are not-null, we also have $x=0$ as (maybe multiple) root, so there will be $n$ total real roots.