Solution:

WLOG $a_{1} > a_{2} > \ldots > a_{n}$. The graph of each $\dfrac{a_{i}}{x + a_{i}}$ is a rectangular hyperbola with asymptotes $x = -a_{i}$ and $y = 0$. So it is not hard to see that the graph of $f(x)$ is made up of $n + 1$ strictly decreasing parts. For $x < -a_{1}$, $f(x)$ is negative. For $x \in (-a_{i}, -a_{i+1})$, $f(x)$ decreases from $\infty$ to $-\infty$. Finally, for $x > -a_{n}$, $f(x)$ decreases from $\infty$ to $0$. Thus $f(x) = 1$ at $n$ values $b_{1} < b_{2} < \ldots < b_{n}$, and $f(x) \geq 1$ on the $n$ intervals $(-a_{1}, b_{1}),\ (-a_{2}, b_{2}),\ \ldots,\ (-a_{n}, b_{n})$. So the sum of the lengths of these intervals is $\sum (a_{i} + b_{i})$. We show that $\sum b_{i} = 0$.

Multiplying $f(x) = 1$ by $\prod (x + a_{j})$ we get a polynomial of degree $n$:
$$\prod (x + a_{j}) - \sum_{i} \left(a_{i} \prod_{j \neq i} (x + a_{j})\right) = 0$$
The coefficient of $x^{n}$ is $1$ and the coefficient of $x^{n-1}$ is $\sum a_{j} - \sum a_{i} = 0$. Hence the sum of the roots, which is $\sum b_{i}$, is zero.

Therefore, the sum of the lengths of the intervals is $\sum a_{i}$.