Austrian Mathematical Olympiad

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:

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

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 $$\begin{array}{rl}2& =23l-11k\\ & =l+11(2l-k).\end{array}$$ We obtain $l\equiv 2(\mathrm{mod}11)$. 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, $N=5$.