The answer is $\boxed{3077493}$ To prove that $C\le 3077493$ you can take $1013$ sticks of $1\times 2025$ and $1013$ sticks of $1\times 1013$. FTSOC assume that all the $1\times 2025$ sticks are placed vertically. Note that the other sticks should also be vertically and it's a contradiction since we have $2025$ columns and each column has at most one stick in it. Thus $C<1013\times 2025+1013\times 1013=3077494$ Now for the algorithm, we can assume that all sticks have length $\ge 1013$ (except for one maybe). Because we can turn two $1\times r$ and $1\times s$ sticks (where $r,s\le 1012$) into one stick of length $r+s$. Now start from the top-right corner and place the sticks horizontally in decreasing order till you reach the bottom row, now go to the bottom-left corner and place the remaining sticks vertically in decreasing order till we can't place another stick which means the stick that we've placed has a common cell with one of those horizontal sticks. Let this cell be the cell $(a,b)$. You can easily notice that so far we have placed $2025-b$ sticks of length at least $2025-a$ and $a+b+1$ sticks of length at least $b+1$ and now we have a problem. So if we run into a problem, then $C\ge (2025-a)(2025-b)+(a+b+1)(b+1)=2025^2+b+1-(a+b)(2024-b)+ab$ so $C\ge b+1+(2025^2-1012^2)+(a+b-1012)^2-a^2\ge 2025^2-1012^2+1013+(b-1012)(b+2a-1012)\ge 3077494$. Hence if we run into a problem in this algorithm, then $C\ge 3077494$. But we have assumed that $C<3077494$ for the algorithm. Thus the answer is $3077493$ and we are done