Let $A_n=\sum_{j=1}^{n}\left[j\sum_{k=1}^{n-j+1} k\right]$. Write out $A_{n-1}$ and notice that $A_n-A_{n-1}=\sum_{j=1}^{n}\sum_{k=1}^{j}k$ (write out terms by terms if necessary). Since $n^4$ is strictly monotone and divergent sequence, by Stolz–Cesàro
$$ \lim _{n \rightarrow \infty} \frac{A_n}{n^4}=\frac{A_n-A_{n-1}}{n^4-(n-1)^4}=\lim _{n \rightarrow \infty} \frac{\sum_{j=1}^{n}\sum_{k=1}^{j}k}{4n^3-6n^2+4n-1}. $$$$ \lim _{n \rightarrow \infty} \frac{A_n}{n^4}=\lim _{n \rightarrow \infty}\frac{A_n-A_{n-1}}{n^4-(n-1)^4}=\lim _{n \rightarrow \infty} \frac{\sum_{j=1}^{n}\sum_{k=1}^{j}k}{4n^3-6n^2+4n-1}. $$ Letting $B_n=\sum_{j=1}^{n}\sum_{k=1}^{j}k$ and applying Stolz–Cesàro similarly again
$$ \lim _{n \rightarrow \infty} \frac{B_n}{4n^3-6n^2+4n-1}=\lim _{n \rightarrow \infty} \frac{\sum_{k=1}^{n}k}{12n^2-24n+14}. $$ And just for consistency $C_n=\sum_{k=1}^{n}k$ and for the last time
$$ \lim _{n \rightarrow \infty} \frac{C_n}{12n^2-24n+14}=\lim_{n\to \infty}\frac{n}{24n-36}=\frac{1}{24}. $$ Hence the answer is $\frac{96}{24}=4$.