All-Soviet-Union Mathematical Olympiad

Solution:
Assume $a_{1} < a_{2} < \ldots < a_{n}$. We have the following collection of increasing sums:

$a_{1} < a_{2} < \ldots < a_{n}$ ($n$ sums)

$a_{1} + a_{n} < a_{2} + a_{n} < \ldots < a_{n - 1} + a_{n}$ ($n-1$ sums)

$a_{1} + a_{n - 1} + a_{n} < a_{2} + a_{n - 1} + a_{n} < \ldots < a_{n - 2} + a_{n - 1} + a_{n}$ ($n-2$ sums)

$\vdots$

$a_{1} + a_{2} + \ldots + a_{n}$ (1 sum)

A total of $1 + 2 + \ldots + n = \frac{n(n + 1)}{2}$.