Find the greatest integer $k \leq 2023$ for which the following holds: whenever Alice colours exactly $k$ numbers of the set $\{1,2, \ldots, 2023\}$ in red, Bob can colour some of the remaining uncoloured numbers in blue, such that the sum of the red numbers is the same as the sum of the blue numbers.
Answer: 592.
For $k \geq 593$, Alice can color the greatest 593 numbers $1431,1432, \ldots, 2023$ and any other $(k-593)$ numbers so that their sum $s$ would satisfy

$$
s \geq \frac{2023 \cdot 2024}{2}-\frac{1430 \cdot 1431}{2}>\frac{1}{2} \cdot\left(\frac{2023 \cdot 2024}{2}\right),
$$

thus anyhow Bob chooses his numbers, the sum of his numbers will be less than Alice's numbers' sum.

We now show that $k=592$ satisfies the condition. Let $s$ be the sum of Alice's 592 numbers; note that $s < \frac{2023 \cdot 2024}{4}$.

Case 1. $s_{0} > 2023$. Note that $s < \frac{2023 \cdot 2024}{4}$, so $s_{0} = \frac{2023 \cdot 2024}{2} - 2s$. Since $s_{0} > 2023$, we have $s < \frac{2023 \cdot 2024}{4} - 1012$. Thus, Alice cannot have colored all the numbers from 1432 to 2023, and some of these pairs have no red-colored components, so Bob can choose one of these pairs and color those two numbers in blue. Thus 594 numbers are colored so far.

Further, the 1011 pairs

$$
(1,2023),(2,2022), \ldots,(1011,1013)
$$

have sum of the components equal to 2024. Among these, at least $1011-594=417 > 337 \geq a$ pairs have no components colored, so Bob can choose $a$ (or $a-1$) uncolored pairs and color them all blue to achieve a collection of blue numbers with their sum equal to $s_{0}$.

Case 2. $s_{0} \leq 2023$. Note that $s \geq 1+2+\ldots+592 > 2023$, thus we have $s_{0} = \frac{2023 \cdot 2024}{2} - 2s$, i.e. $s = \frac{2023 \cdot 2024}{4} - \frac{s_{0}}{2}$.

If $s_{0} > 2 \cdot 593$, at least one of the 593 pairs

$$
\left(1, s_{0}-1\right),\left(2, s_{0}-2\right), \ldots,\left(593, s_{0}-593\right)
$$

have no red-colored components, so Bob can choose these two numbers and immediately achieve the sum of $s_{0}$. And if $s_{0} \leq 2 \cdot 593$, then
$s = \frac{2023 \cdot 2024}{4} - \frac{s_{0}}{2} \geq (1432+1433+\ldots+2023) - 593 = 839 + (1434+1435+\ldots+2023)$,
hence Alice cannot have colored any of the numbers $1,2, \ldots, 838$. Then Bob can easily choose one or two of these numbers having the sum of $s_{0}$.