Given any prime $p$ and positive integer $x$, let $v_{p}(x)$ denote the highest power of $p$ dividing $x$. We claim that Jane cannot write 2024 such numbers as that would imply that $1! \cdot 2! \cdots 2024!$ is the square of the product of the 2024 numbers. Let $p$ be a prime and $k$ be a natural number such that $kp < 2024$. Then note that

$$v_{p}(1! \cdot 2! \cdots 2024!) = (2024 - p + 1) + (2024 - 2p + 1) + \ldots + (2024 - kp + 1)$$

In particular, let $p$ be in $\left(\frac{2024}{4}, \frac{2024}{2}\right)$. By Bertrand's Postulate, such a prime $p$ exists (and $p$ must also be odd). Further, the corresponding $k$ is either 2 or 3. Either way, $v_{p}(1! \cdot 2! \cdots 2024!)$ is odd from the above formula, and so $1! \cdot 2! \cdots 2024!$ cannot be a perfect square.

1. Let the numbers on the circle be $a_1, a_2, \ldots, a_{2024}$. Suppose $b_i!$ are a permutation of $1!, 2!, \ldots, 2024!$ for $i = 1, 2, \ldots, 2024$. If this is possible, then $a_j a_{j+1} = b_j!$ where $a_1 = a_{2025}$.

2. Consider the product of all the adjacent pairs:
$$b_1! b_2! \cdots b_{2024}! = (a_1 a_2) (a_2 a_3) \cdots (a_{2023} a_{2024}) (a_{2024} a_1)$$
This simplifies to:
$$(a_1 a_2 a_3 \cdots a_{2024})^2$$
Let $P = a_1 a_2 \cdots a_{2024}$. Then:
$$b_1! b_2! \cdots b_{2024}! = P^2$$

3. Let $\tau(n)$ be the total number of positive divisors of $n$. We have:
$$\tau(b_1! b_2! \cdots b_{2024}!) = \tau(P^2)$$
Since $P^2$ is a perfect square, $\tau(P^2)$ is always odd.

4. Consider the prime number $661$. It divides:
$$b_1! b_2! \cdots b_{2024}! = 1! 2! \cdots 2024! = C$$
We need to find the exponent of $661$ in $C$. Using Legendre's formula for the exponent of a prime $p$ in $n!$:
$$v_{661}(C) = \left\lfloor \frac{2024}{661} \right\rfloor + \left\lfloor \frac{2024}{661^2} \right\rfloor + \left\lfloor \frac{2024}{661^3} \right\rfloor + \cdots$$
Calculating each term:
$$\left\lfloor \frac{2024}{661} \right\rfloor = 3, \quad \left\lfloor \frac{2024}{661^2} \right\rfloor = 0 \quad (\text{since } 661^2 > 2024)$$
Thus:
$$v_{661}(C) = 3$$

5. The number of divisors of $C$ is given by:
$$\tau(C) = \tau(1! 2! \cdots 2024!)$$
Since $v_{661}(C) = 3$, the factor $661^3$ contributes $3+1 = 4$ to $\tau(C)$. Therefore, $\tau(C)$ is even.

6. We have a contradiction because $\tau(C)$ must be even, but $\tau(P^2)$ is odd. Therefore, Jane cannot accomplish her goal.

$\blacksquare$