Trick to change the bounds for the summations $$\sum_{i=1}^{N-1}\sum_{j=i+1}^{N} |A_i - A_j| $$
$$= \frac{1}{2}\sum_{i=1}^{N-1}\sum_{j=i+1}^{N} 2|A_i - A_j| $$
$$= \frac{1}{2}\sum_{i=1}^{N-1}\sum_{j=i+1}^{N}|A_i - A_j| + |A_j - A_i|$$
making it symmetric and can write that $$\frac{1}{2}\sum_{i=1}^{N-1}\sum_{j=i+1}^{N}|A_i - A_j| + |A_j - A_i| $$ $$=  \frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}|A_i - A_j|$$ My question is, is there a formula based expression that enables you to make the sums of i and j with the same bounds? I'm looking for expressive details into what happened here to get that $$\frac{1}{2}\sum_{i=1}^{N-1}\sum_{j=i+1}^{N}|A_i - A_j| + |A_j - A_i|$$ $$ =  \frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}|A_i - A_j|$$
 A: It really is very easy to see if you follow Benjamin Wang’s suggestion in the comments to draw an $N\times N$ square with $|A_i-A_j|$ in position $\langle i,j\rangle$. However, here is an algebraic demonstration.
First note that
$$\begin{align*}
&\sum_{i=1}^{N-1}\sum_{j=i+1}^N\big(|A_i - A_j|+|A_j - A_i|\big)\\
&\qquad=\sum_{1\le i<j\le N}\big(|A_i - A_j|+|A_j - A_i|\big)\\
&\qquad=\sum_{1\le i<j\le N}|A_i-A_j|+\sum_{1\le\color{red}i<\color{blue}j\le n}|\color{blue}{A_j}-\color{red}{A_i}|\\
&\qquad=\sum_{1\le i<j\le N}|A_i-A_j|+\sum_{1\le\color{red}j<\color{blue}i\le n}|\color{blue}{A_i}-\color{red}{A_j}|\,.\tag{1}
\end{align*}$$
In the last step I simply interchanged the names $i$ and $j$ in the second summation. The in each line blue symbols go with the larger index, whatever it’s called, and the red symbols go with the smaller index.
The first summation in $(1)$ covers all pairs $\langle i,j\rangle$ with $i<j$; the second, all pairs $\langle i,j\rangle$ with $i>j$. The pairs $|A_i-A_j|$ with $i=j$ are of course all $0$, so $(1)$ is actually the sum of the quantities $|A_i-A_j|$ over all pairs $\langle i,j\rangle$ with $1\le i,j\le n$, which is
$$\sum_{i=1}^N\sum_{j=1}^N|A_i-A_j|\,.$$
Added: I’m adding on example of the visual approach suggested by Benjamin Wang.
I’ll take $N=6$ for my example. For $1\le i,j\le 6$ let $a_{ij}=|A_i-A_j|$. Now
$$\begin{align*}
\sum_{i=1}^5\sum_{j=i+1}^6\left(|A_i-A_j|+|A_j-A_i|\right)&=\sum_{i=1}^5\sum_{j=i+1}^6(a_{ij}+a_{ji})\\
&=\color{red}{\sum_{i=1}^5\sum_{j=i+1}^6a_{ij}}+\color{blue}{\sum_{i=1}^5\sum_{j=i+1}^6a_{ji}}\,.
\end{align*}$$
In the array below I’ve used the red and blue colors to pick out the terms in those last two summations.
$$\begin{array}{ccc}
a_{11}&\color{red}{a_{12}}&\color{red}{a_{13}}&\color{red}{a_{14}}&\color{red}{a_{15}}&\color{red}{a_{16}}\\
\color{blue}{a_{21}}&a_{22}&\color{red}{a_{23}}&\color{red}{a_{24}}&\color{red}{a_{25}}&\color{red}{a_{26}}\\
\color{blue}{a_{31}}&\color{blue}{a_{32}}&a_{33}&\color{red}{a_{34}}&\color{red}{a_{35}}&\color{red}{a_{36}}\\
\color{blue}{a_{41}}&\color{blue}{a_{42}}&\color{blue}{a_{43}}&a_{44}&\color{red}{a_{45}}&\color{red}{a_{46}}\\
\color{blue}{a_{51}}&\color{blue}{a_{52}}&\color{blue}{a_{53}}&\color{blue}{a_{54}}&a_{55}&\color{red}{a_{56}}\\
\color{blue}{a_{61}}&\color{blue}{a_{62}}&\color{blue}{a_{63}}&\color{blue}{a_{64}}&\color{blue}{a_{65}}&a_{66}\\
\end{array}$$
As you can see, we’re adding up all of the red and blue entries in the array, omitting only the black entries on the diagonal. But $a_{ii}=|A_i-A_i|=0$ for all $i$, so the black entries are all $0$:
$$\begin{array}{ccc}
0&\color{red}{a_{12}}&\color{red}{a_{13}}&\color{red}{a_{14}}&\color{red}{a_{15}}&\color{red}{a_{16}}\\
\color{blue}{a_{21}}&0&\color{red}{a_{23}}&\color{red}{a_{24}}&\color{red}{a_{25}}&\color{red}{a_{26}}\\
\color{blue}{a_{31}}&\color{blue}{a_{32}}&0&\color{red}{a_{34}}&\color{red}{a_{35}}&\color{red}{a_{36}}\\
\color{blue}{a_{41}}&\color{blue}{a_{42}}&\color{blue}{a_{43}}&0&\color{red}{a_{45}}&\color{red}{a_{46}}\\
\color{blue}{a_{51}}&\color{blue}{a_{52}}&\color{blue}{a_{53}}&\color{blue}{a_{54}}&0&\color{red}{a_{56}}\\
\color{blue}{a_{61}}&\color{blue}{a_{62}}&\color{blue}{a_{63}}&\color{blue}{a_{64}}&\color{blue}{a_{65}}&0\\
\end{array}$$
Thus, we can get the same sum by adding up all of the entries in the entire square array, and that sum is
$$\sum_{i=1}^6\sum_{j=1}^6a_{ij}\,.$$
In other words,
$$\sum_{i=1}^5\sum_{j=i+1}^6(a_{ij}+a_{ji})=\sum_{i=1}^6\sum_{j=1}^6a_{ij}\,,$$
i.e.,
$$\sum_{i=1}^5\sum_{j=i+1}^6\left(|A_i-A_j|+|A_j-A_i|\right)=\sum_{i=1}^6\sum_{j=1}^6\left(|A_i-A_j|+|A_j-A_i|\right)\,.$$
And the same argument works for any $N$, not just $6$.
