Horrible combinatoric summation

$\begin{aligned} A&=\sum_{k=0}^{n-1}\sum_{m=0}^k\sum_{l=k+1}^n \binom nm \binom nl\\ &=\sum_{l=1}^n \sum_{m=0}^{l-1}\sum_{k=m}^{l-1} \binom nm \binom nl &&\small (0\;\leqslant\; m\;\leqslant\; k\;\leqslant\; l-1\;\leqslant\; n-1)\\ &=\sum_{l=1}^n \sum_{m=0}^{l-1} (l-m)\binom nm \binom nl\\ &=\sum_{l=1}^n \sum_{r=1}^l r\binom n{l-r} \binom nl &&\small(r=l-m)\\ &=\sum_{r=1}^n r\sum_{l=r}^n\binom n{l-r} \binom n{n-l}\\ &=\sum_{r=1}^n r\binom {2n}{n-r} &&\small(\text{Vandermonde})\\ &=\frac n2 \binom {2n}n &&\small(\text{see Note})\\\\ B&=\sum_{k=0}^n\binom{n}k^2 \\ &=\sum_{k=0}^n\binom{n}k\binom {n}{n-k} \\ &=\binom {2n}n &&\small(\text{Vandermonde})\\\\ \therefore \frac AB &= \frac n2 =\color{#D61F06}{1010} &&\small(n=2020) \end{aligned}$

Note:
$\small \begin{aligned} &\qquad \sum_{r=1}^n r\binom {2n}{n-r} \color{grey}{=\sum_{r=1}^n r\binom {2n}{n+r}}\\ &=\frac 12 \sum_{r=1}^n (n+r)\binom {2n}{n-r}-(n-r)\binom {2n}{n+r}\\ &=\frac 12 \sum_{r=1}^n 2n\binom {2n-1}{n-r}-2n\binom {2n-1}{n+r}\\ &=\frac {2n}2 \sum_{r=1}^n \binom {2n-1}{n-r}-\sum_{r=1}^{n-1} \binom {2n-1}{n-r-1}\\ &=\frac {2n}2\left[\sum_{r=1}^n \binom {2n-1}{n-r} - \sum_{r=2}^n \binom {2n-1}{n-r} \right]\\ &=\frac {2n}2\binom {2n-1}{n-1}\\ &=\frac n2 \binom {2n}n\\ \end{aligned}$

Let $N \in \mathbb{N}$ . For any $0 \le u \le N$ , the sum $C_{N,u} \; = \; \sum_{m=0}^u \binom{N}{m}\binom{N}{u-m}$ is the coefficient of $X^u$ in the polynomial $(1+X)^N \times (1+X)^N \; = \; (1 + X)^{2N}$ , and hence $C_{N,u} \; =\; \binom{2N}{u} \hspace{2cm} 0 \le u \le N$ In particular we see that $B_N \; = \; \sum_{k=0}^N \binom{N}{k}^2 \; = \; \sum_{k=0}^N \binom{N}{k}\binom{N}{N-k} \; = \; C_{N,N} \; = \; \binom{2N}{N}$ On the other hand $\begin{aligned} A_N & = \; \sum_{k=0}^{N-1} \left(\sum_{m=0}^k \binom{N}{m}\right)\left(\sum_{\ell=k+1}^N\binom{N}{\ell}\right) \; = \; \sum_{m=0}^{N-1}\sum_{\ell=m+1}^N \sum_{k=m}^{\ell-1}\binom{N}{m}\binom{N}{\ell} \; = \; \sum_{m=0}^{N-1}\sum_{\ell=m+1}^N (\ell - m)\binom{N}{m}\binom{N}{\ell} \\ & = \; \sum_{m=0}^{N-1}\sum_{\ell=m+1}^N (\ell - m)\binom{N}{m}\binom{N}{N - \ell} \; = \; \sum_{m=0}^{N-1}\sum_{\ell=0}^{N-1-m} (N - m - \ell)\binom{N}{m}\binom{N}{\ell} \; = \; \sum_{u=0}^{N-1} (N-u)\sum_{m=0}^u \binom{N}{m}\binom{N}{u-m} \\ & = \; \sum_{u=0}^{N-1}(N-u)C_{N,u} \; = \; \sum_{u=0}^{N-1}(N-u)\binom{2N}{u} \; = \; \frac12\sum_{u=0}^{N-1}\big((2N-u) - u\big)\binom{2N}{u} \\ &= \; \frac12\left[\sum_{u=0}^{N-1} \frac{(2N)!}{u! (2N-1-u)!} - \sum_{u=1}^{N-1} \frac{(2N)!}{(u-1)!(2N-u)!}\right] \; = \; \frac12\left[\sum_{u=0}^{N-1} \frac{(2N)!}{u! (2N-1-u)!} - \sum_{u=0}^{N-2} \frac{(2N)!}{u!(2N-1-u)!}\right] \; = \; \frac{(2N)!}{2(N-1)! N!} \\ & = \; \frac12N\binom{2N}{N} \end{aligned}$ so that $\frac{A_N}{B_N} \; = \; \frac12N$ In the case $N=2020$ , we obtain the answer $\boxed{1010}$ .