$\displaystyle f(x)=\bigg(\sum\limits_{i=1}^n\frac{a_i}{a_i-x}\bigg)-2015\quad$ with all $a_i>0$.
$\displaystyle f'(x)=\sum\limits_{i=1}^n\underbrace{\frac{a_i}{(a_i-x)^2}}_{>0}>0\quad$ so $f$ is increasing on each interval where it is defined.
It is also clear that $f$ is continuous except in its poles, thus on $\mathbb R\setminus\{a_1,a_2,...,a_n\}$.
$\begin{cases}
\displaystyle \lim\limits_{x\to a_i^-}f(x)=+\infty\\
\displaystyle \lim\limits_{x\to a_i^+}f(x)=-\infty\\
\end{cases}$
By applying intermediate value theorem and using that $f\ \nearrow$ strictly, there is exactly $1$ real root in $]a_i,a_{i+1}[$, making this $n-1$ roots for $i=1..n$.
We have also $\lim\limits_{x\to-\infty}f(x)=-2015<0$ so there is $1$ root in $]-\infty,a_1[$.
We have also $\lim\limits_{x\to+\infty}f(x)=-2015<0$ so there is no root in $]a_n,+\infty[$.
Combining the results, gives exactly $n$ roots in $\mathbb R$.