Below shows how to manipulate the double sum into a single sum, apologies if it contains too much detail but it was difficult to explain without actually doing it.
To move from the first line to the second, in the left square bracket the inner sum is over $j$ but the index is $i$ consequently this just adds up the same term $n-i$ times. In the square bracket on the right is the non trivial part, to realise this relationship you can expand the terms by trying values for $n$ and then spot the pattern of how often each term appears.
To move from the second line to the third, arrange the left square bracket so that the summation limits match the right square bracket by pulling out unwanted terms, then combine the sums in the fourth line; you will then see that you can stick back in the loose term to form the fifth line.
$$\begin{aligned}
\sum _{i=1}^{n-1} \left( \sum _{j=i+1}^{n}\epsilon_{{i}}-\epsilon_{{j} } \right) &=\left[\sum_{i=1}^{n-1} \left( \sum _{j=i+1}^{n}\epsilon_{{i}} \right) \right]-\left[\sum_{i=1}^{n-1} \left( \sum_{j=i+1}^{n}\epsilon_{{j}} \right)\right] \\
&=\left[\sum _{i=1}^{n-1} \left( n-i \right)\epsilon_{{i}}\right]-\left[\sum _{ i=2}^{n} \left( i-1 \right)\epsilon_{{i}}\right]\\
&=\left[\left( n-1 \right) \epsilon_{{1}}+\sum _{i=2}^{n} \left( n-i \right) \epsilon_{{i}}\right]-\left[\sum _{i=2}^{n} \left( i-1 \right) \epsilon_{{i}}\right]\\ &=\left( n-1 \right) \epsilon_{{1}}+\sum _{i=2}^{n} \left( n-2\,i+1 \right) \epsilon_{{i}}\\
&=\sum _{i=1}^{n} \left( n-2\,i+1 \right) \epsilon_{{i}}\\
\end{aligned}$$
