We observe that $2024 = 2^3 \cdot 11 \cdot 23$. An integer has a common divisor greater than $1$ with $2024$ if and only if it is divisible by $2$, $11$ or $23$.
Let $N$ be the desired maximal number. Only each $11$th integer is divisible by $11$. That means that if $z$ is divisible by $11$, then $z+1, z+2, \dots, z+10$ are not divisible by $11$. Analogously, $23$ divides only every $23$rd integer. Six consecutive integers contain exactly three odd numbers. At most one of them is divisible by $11$ and at most one of them is divisible by $23$. This shows that $N \le 5$.

Now, we try to find five consecutive integers $n, n+1, n+2, n+3, n+4$ that have a common divisor greater than $1$ with $2024$.
We can do that in the following way:

$$\begin{array}{c|l} n & \text{even} \\ n + 1 & \text{divisible by } 11 \\ n + 2 & \text{even} \\ n + 3 & \text{divisible by } 23 \\ n + 4 & \text{even} \end{array}$$

That means that we want $n + 1 = 11k$ and $n + 3 = 23l$ with $k$ and $l$ odd. If we subtract the second equation from the first, we get
$$\begin{aligned} 2 &= 23l - 11k \\ &= l + 11(2l - k). \end{aligned}$$
We obtain $l \equiv 2 \pmod{11}$. We see that $l = 13$ works, since we get $n + 3 = 23l = 299$ and therefore $n = 296$ and the five consecutive integers $296, 297, 298, 299, 300$, which have the desired property.
Therefore, $N = 5$.