I claim it is impossible. Assume the contrary, i.e. you can make a square $k \times k$ using these rectangles for some $k\ge 2$. Obviously, we cannot use rectangles $1\times m$ for $m>k$. So we can only use rectangles of the form $1\times l$ for $l\le k$. However, the number of "tiles" we can fill using some (possibly all) rectangles of the form $1\times l$ for $l\le k$ is at most $1+2+\dots+k=\frac{k(k+1)}{2}.$ However, we want these rectangles to fill up a square $k\times k$, i.e. $k^2$ "tiles". Thus, we must have $$1+2+\dots+k=\frac{k(k+1)}{2}\ge k^2 \iff 1\ge k,$$which is impossible. Hence, we cannot construct a square with side length greater than $1$.

We assume that there is a solution on posoitive integer $T>1$ we see that 1-we can't put an rectangle with lenght $T+x$ in the square 2-the sum of areas in rectangles that is length is less or equal to $T$ is $1*1+2*1+\dots +1*T=1+2+\dots +T=\cfrac {T(T+1)}{2}$ from assuming we get than the area of square is less or equal to the sum of areas $\implies \cfrac {T(T+1)}{2} \ge T^2 \implies 1 \ge T$ wrong from asumming so there is no solution

A $k\times k$ square can only fit rectangles equal or smaller to a $1\times k$ but this is insufficient to cover the area.

it is impossible. firstly if we can make square with lenght K , K can not be larger then 2024, because we only can "cover" $\cfrac{2024\times2023}{2}$ area and it is not enough for 2024*2024( square with side lenghts 2024) then if K<2024. we need to cover K^2 area with rectangles, but we can only use rectangles with side lenght <K and sum of areas of this rectangles are (1+k)*k/2 with is less then K^2 so we can not make square