Given the positive real numbers $a_{1}<a_{2}<\cdots<a_{n}$, consider the function \[f(x)=\frac{a_{1}}{x+a_{1}}+\frac{a_{2}}{x+a_{2}}+\cdots+\frac{a_{n}}{x+a_{n}}\] Determine the sum of the lengths of the disjoint intervals formed by all the values of $x$ such that $f(x)>1$.
1. Given the function 
   \[
   f(x) = \frac{a_1}{x + a_1} + \frac{a_2}{x + a_2} + \cdots + \frac{a_n}{x + a_n}
   \]
   we need to determine the sum of the lengths of the disjoint intervals formed by all the values of \( x \) such that \( f(x) > 1 \).

2. Let us rewrite \( f(x) \) in the form of a single fraction:
   \[
   f(x) = \frac{P(x)}{Q(x)}
   \]
   where 
   \[
   P(x) = \sum_{i=1}^n a_i \prod_{j \neq i} (x + a_j)
   \]
   and 
   \[
   Q(x) = \prod_{i=1}^n (x + a_i).
   \]

3. We need to find the values of \( x \) for which \( f(x) > 1 \). This is equivalent to solving:
   \[
   \frac{P(x)}{Q(x)} > 1 \quad \Leftrightarrow \quad P(x) > Q(x) - P(x).
   \]

4. Simplifying the inequality:
   \[
   P(x) > Q(x) - P(x) \quad \Leftrightarrow \quad 2P(x) > Q(x) \quad \Leftrightarrow \quad P(x) - \frac{Q(x)}{2} > 0.
   \]

5. To find the roots of the equation \( f(x) = 1 \), we solve:
   \[
   \frac{P(x)}{Q(x)} = 1 \quad \Leftrightarrow \quad P(x) = Q(x) - P(x) \quad \Leftrightarrow \quad 2P(x) = Q(x).
   \]

6. The polynomial \( Q(x) - P(x) \) has exactly \( n \) roots, denoted as \( r_1, r_2, \ldots, r_n \), where:
   \[
   -a_n < r_n < -a_{n-1} < r_{n-1} < \cdots < -a_1 < r_1.
   \]

7. The sum of the lengths of the disjoint intervals where \( f(x) > 1 \) is given by:
   \[
   \sum_{i=1}^n (r_i + a_i).
   \]

8. To find \( \sum_{i=1}^n r_i \), we note that it is the opposite of the coefficient of \( x^{n-1} \) in \( Q(x) - P(x) \).

9. The coefficient of \( x^{n-1} \) in \( Q(x) \) is:
   \[
   \sum_{i=1}^n a_i.
   \]

10. The coefficient of \( x^{n-1} \) in \( P(x) \) is also:
    \[
    \sum_{i=1}^n a_i.
    \]

11. Therefore, the coefficient of \( x^{n-1} \) in \( Q(x) - P(x) \) is:
    \[
    \sum_{i=1}^n a_i - \sum_{i=1}^n a_i = 0.
    \]

12. Hence, \( \sum_{i=1}^n r_i = 0 \).

13. The sum of the lengths of the disjoint intervals is:
    \[
    \sum_{i=1}^n (r_i + a_i) = \sum_{i=1}^n r_i + \sum_{i=1}^n a_i = 0 + \sum_{i=1}^n a_i = \sum_{i=1}^n a_i.
    \]

The final answer is \(\boxed{\sum_{i=1}^n a_i}\).