So here is my problem. If we have a vector $\textbf{x}=(x_1,...,x_n)$ where $x_j \in \mathbb{N}$ for each $j \in \{1,...,n\}$, then is there a way to maximize the value of the following combinatorial identity:
$\sum_{j=1}^{n}\binom{2x_j}{2}$?
The problem here is that we do not know anything about the vector $\textbf{x}$ except for the fact that the entires are all natural numbers and that the constraint is $\sum_{j=1}^{n}x_{j}=k$ where $k\in \mathbb{N}$ is fixed.
On
Note that whenever $x$ and $y$ are positive integers, $$ \binom{2(x+y)}2> \binom{2x}2+\binom{2y}2 $$ Therefore, if you have a vector $(x_1,\dots,x_n)$ where there are two nonzero components, then you get a larger value of $\sum_i \binom{2x_i}2$ by combining those two nonzero components into one. This means that the vector which attains the optimum must have only one nonzero component, so it is some permutation of $(k,0,0,\dots,0)$.
You want one component of $\bf x$ to have value $k$ and all the rest to be $0$. Because the term ${2x_j \choose 2}=x_j(2x_j-1)$ you will increase the sum every time you move $1$ to the largest component from any other component. This stops when one component equals $k$.