The sequence $\{a_n\}_{n\ge 1}$ of real numbers is defined as follows:
$$a_1=1, \quad \text{and}\quad a_{n+1}=\frac{1}{2\lfloor a_n \rfloor -a_n+1} \quad \text{for all} \quad n\ge 1$$
Find $a_{2024}$.
1. We start with the given sequence $\{a_n\}_{n \ge 1}$ defined by:
   \[
   a_1 = 1, \quad \text{and} \quad a_{n+1} = \frac{1}{2\lfloor a_n \rfloor - a_n + 1} \quad \text{for all} \quad n \ge 1
   \]

2. To analyze the sequence, we write $a_n$ in the form of a fraction $\frac{p_n}{q_n}$ in its reduced form. This means $p_n$ and $q_n$ are coprime integers.

3. Given $a_n = \frac{p_n}{q_n}$, we substitute this into the recurrence relation:
   \[
   a_{n+1} = \frac{1}{2\lfloor a_n \rfloor - a_n + 1}
   \]
   Since $a_n = \frac{p_n}{q_n}$, we have:
   \[
   \lfloor a_n \rfloor = \left\lfloor \frac{p_n}{q_n} \right\rfloor
   \]
   Therefore:
   \[
   a_{n+1} = \frac{1}{2\left\lfloor \frac{p_n}{q_n} \right\rfloor - \frac{p_n}{q_n} + 1}
   \]

4. To simplify, we multiply the numerator and the denominator by $q_n$:
   \[
   a_{n+1} = \frac{q_n}{q_n \left(2\left\lfloor \frac{p_n}{q_n} \right\rfloor + 1\right) - p_n}
   \]
   This form is still in reduced form, so we can identify:
   \[
   p_{n+1} = q_n \quad \text{and} \quad q_{n+1} = q_n \left(2\left\lfloor \frac{p_n}{q_n} \right\rfloor + 1\right) - p_n
   \]

5. From the recurrence relation, we observe that $a_n = \frac{p_n}{p_{n+1}}$ where $p_1 = p_2 = 1$. The sequence $p_{n+1} = p_n \left(2\left\lfloor \frac{p_{n-1}}{p_n} \right\rfloor + 1\right) - p_{n-1}$ is known as the Stern-Brocot sequence.

6. The Stern-Brocot sequence is a well-known sequence where every positive rational number appears exactly once. This sequence is indexed in the OEIS as A002487.

7. To find $a_{2024}$, we note that the sequence $a_n$ is periodic with period 3, as it cycles through the values 1, 1/2, and 2. This can be verified by computing the first few terms:
   \[
   a_1 = 1
   \]
   \[
   a_2 = \frac{1}{2\lfloor 1 \rfloor - 1 + 1} = \frac{1}{2 \cdot 1 - 1 + 1} = \frac{1}{2}
   \]
   \[
   a_3 = \frac{1}{2\lfloor \frac{1}{2} \rfloor - \frac{1}{2} + 1} = \frac{1}{2 \cdot 0 - \frac{1}{2} + 1} = 2
   \]
   \[
   a_4 = \frac{1}{2\lfloor 2 \rfloor - 2 + 1} = \frac{1}{2 \cdot 2 - 2 + 1} = 1
   \]
   Thus, the sequence repeats every 3 terms: $1, \frac{1}{2}, 2$.

8. Since the sequence is periodic with period 3, we find the position of $a_{2024}$ in the cycle:
   \[
   2024 \mod 3 = 2
   \]
   Therefore, $a_{2024}$ corresponds to the second term in the cycle, which is $\frac{1}{2}$.

The final answer is $\boxed{\frac{1}{2}}$