, 2025

Solution:
Define the bounding box of a polyomino to be the smallest axis-aligned rectangle that contains the entire polyomino. Suppose a polyomino satisfying the given conditions has a bounding box with dimensions $w \times h$.

Claim 1. $w + h \leq 90$.

Proof. The polyomino has at least $2w$ horizontal edges and at least $2h$ vertical edges. Moreover, it has a perimeter of $180$. Therefore, $2w + 2h \leq 180$, so $w + h \leq 90$. $\square$

Claim 2. The dimensions of the bounding box are either $44 \times 46$, $45 \times 45$, or $46 \times 44$.

Proof. Note that $hw \geq 2024$ since it contains the polyomino with area $2024$. Suppose for sake of contradiction that $h + w \leq 89$. Then,
$$(h - w)^2 = (h + w)^2 - 4hw \leq 89^2 - 4 \cdot 2024 = -175,$$
contradiction. Therefore, $h + w = 90$, so we can let $(h,w) = (45 + x,45 - x)$. Then, $2025 - x^2 = hw \geq 2024$ implies that $x \in \{-1,0,1\}$, as desired. $\square$

In the first and third cases, the bounding box has area $2024$, so it must be the entire polyomino, giving us the $44 \times 46$ rectangle (and its rotation) as a possible answer. In the second case, the bounding box has area $2025$, so one cell must be removed to form the polyomino. Removing the corner cell yields a polyomino with perimeter $180$, and removing any other cells yields a polyomino with perimeter greater than $180$. Therefore, the only other possibility is a $45 \times 45$ square missing a corner. Thus the answer is $\boxed{2}$.