(a) Let \(a_{0}, a_{1}, \ldots, a_{2024}\) be real numbers such that \(\left|a_{i+1}-a_{i}\right| \leqslant 1\) for \(i=0,1, \ldots, 2023\). Find the minimum possible value of 
\[ a_{0} a_{1}+a_{1} a_{2}+\cdots+a_{2023} a_{2024} \]

(b) Does there exist a real number \(C\) such that 
\[ a_{0} a_{1}-a_{1} a_{2}+a_{2} a_{3}-a_{3} a_{4}+\cdots+a_{2022} a_{2023}-a_{2023} a_{2024} \geqslant C \]
for all real numbers \(a_{0}, a_{1}, \ldots, a_{2024}\) such that \(\left|a_{i+1}-a_{i}\right| \leqslant 1\) for \(i=0,1, \ldots, 2023\)?
(a) The minimum value is -506. Note that from $\left|a_{i}-a_{i-1}\right| \leq 1$ it follows that

$$
a_{i} a_{i-1}=\frac{\left(a_{i}+a_{i-1}\right)^{2}-\left(a_{i}-a_{i-1}\right)^{2}}{4} \geq-\frac{\left(a_{i}-a_{i-1}\right)^{2}}{4} \geq-\frac{1}{4}
$$

Adding this for $i=1,2, \ldots, 2024$, we obtain that

$$
a_{0} a_{1}+a_{1} a_{2}+a_{2} a_{3}+\ldots+a_{2023} a_{2024} \geq 2024 \cdot-\frac{1}{4}=-506
$$

We now show that this value can be attained. Indeed, for the sequence $\left(a_{0}, a_{1}, \ldots, a_{2024}\right)=$ $\left(\frac{1}{2},-\frac{1}{2}, \frac{1}{2},-\frac{1}{2}, \frac{1}{2}, \ldots, \frac{1}{2}\right)$ with alternating $\frac{1}{2}$ 's and $-\frac{1}{2}$ 's, each term $a_{i} a_{i-1}$ is equal to $-\frac{1}{4}$, leading to $a_{0} a_{1}+a_{1} a_{2}+a_{2} a_{3}+\ldots+a_{2023} a_{2024}=2024 \cdot-\frac{1}{4}=-506$.
(b) No, such a $C$ does not exist. We argue by contradiction. Suppose $C$ has this property, and consider the sequence defined by $a_{0}=C$ and $a_{i}=C-1$ for $i=1,2, \ldots, 2024$ satisfies the condition in the problem. For this sequence, we have $a_{i} a_{i+1}-a_{i+1} a_{i+2}=0$ for $i=2,4, \ldots$, 2022, so the sum

$$
a_{0} a_{1}-a_{1} a_{2}+a_{2} a_{3}-a_{3} a_{4}+a_{4} a_{5}-a_{5} a_{6}+\ldots+a_{2022} a_{2023}-a_{2023} a_{2024}
$$

is equal to

$$
a_{0} a_{1}-a_{1} a_{2}=C(C-1)-(C-1)^{2}=C-1<C,
$$

contradiction.