Trying to understand a double summation formula

$$ \begin{array}{c} n=3 \\[8pt] \begin{array}{|c|c|ccccccc|c|} % {}_{\large\ell} \, \diagdown \, {}^{\large k} \hline & k & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \ell & (\ell,k)\\ \hline 1 & & (1,1) & (1,2) & (1,3) & (1,4) & & & & \sum_{k=1}^{n+1} \\ 2 & & & (2,2) & (2,3) & (2,4) & (2,5) & & & \sum_{k=2}^{n+2} \\ 3 & & & & (3,3) & (3,4) & (3,5) & (3,6) & & \sum_{k=3}^{n+3} \\ \hline & & \sum_{\ell=1}^1 & \sum_{\ell=1}^2 & \sum_{\ell=1}^3 & \sum_{\ell=4-3}^3 & \sum_{\ell=4-2}^3 & \sum_{\ell=4-1}^3 \\ \hline \end{array} \end{array} $$

One way to handle these changes of variables in a more formal way is to re-write the summation in terms of indicator variables on unrestricted sums, so that interchanging sums is as simple as switching the order. Then we need to manipulate the inequalities defining the indicators until we get the standard form of a sum.

Define $1_P$ to be $1$ when the proposition $P$ is true, otherwise $0$. If we're given variables $a_{l,k}$ with $1 \leq l,k \leq 2n$, we extend $a_{l,k}$ to be $0$ for $l,k$ outside of this range, and use $\sum_l$ to mean $\sum_{l=-\infty}^\infty$. Then we have

\begin{align*} \sum_{l = 1}^n\;\; \sum_{k = l}^{n+l} a_{l,k} &= \sum_l \sum_{k} 1_{1 \leq l \leq n, \, l \leq k \leq n+l} a_{l,k}\\ &= \sum_k \sum_{l} 1_{1 \leq l \leq n, \, l \leq k \leq n+l} a_{l,k}\\ &= \sum_{k=1}^{2n} \sum_{l} 1_{1 \leq l \leq n, \, l \leq k, k \leq n+l} a_{l,k}\\ &= \sum_{k=1}^{2n} \sum_{l} 1_{1 \leq l \leq n, \, l \leq k, k-n \leq l } a_{l,k}\\ &= \sum_{k=1}^{2n} \sum_{l} 1_{1 \leq l \leq n, \, k-n \leq l \leq k} a_{l,k}\\ &= \sum_{k=1}^{2n} \sum_{l=\max(k-n,1)}^{\min(k,n)} a_{l,k}\\ &= \sum_{k=1}^{n} \sum_{l=\max(k-n,1)}^{\min(k,n)} a_{l,k} + \sum_{k=n+1}^{2n} \sum_{l=\max(k-n,1)}^{\min(k,n)} a_{l,k} \\ &= \sum_{k=1}^{n} \sum_{l=1}^{k} a_{l,k} + \sum_{k=n+1}^{2n} \sum_{l=k-n}^{n} a_{l,k}. \\ \end{align*}