We will first prove that for every natural number $n$, there exists at least one natural number between $n$ and $n+17$ that is coprime with $n(n+17)$.
In the case where $n$ is an even number, the required property is satisfied by the number $n+1$. Of course, the numbers $n$ and $n+1$ are coprime. If a number $d > 1$ were a common divisor of $n+1$ and $n+17$, then it would divide the difference $(n+17)-(n+1) = 16$, and thus would be an even number. However, since $n+1$ is an odd number, such a $d$ does not exist. Therefore, the numbers $n+1$ and $n(n+17)$ are coprime.
In the case where $n$ is an odd number, the required property is satisfied by the number $n + 16$. Indeed, the numbers $n+16$ and $n+17$ are coprime. Similarly, as above, we conclude that $n$ and $n+16$ are also coprime. Therefore, $n+16$ and $n(n+17)$ are coprime.
In this way, we have shown that $k \geq 1$.
Consider the number $n = 16!$. The numbers $n+2, n+3, \ldots, n+16$ are not coprime with $n(n+17)$, because for $j = 2, 3, \ldots, 16$, the numbers $n+j$ and $n$ are divisible by $j$. Only the number $n+1$ is coprime with $n(n+17)$. From this example, it follows that $k \leq 1$. Therefore, $k = 1$.
XXXVI OM - III - Problem 1

Determine the largest number $ k $ such that for every natural number $ n $ there are at least $ k $ natural numbers greater than $ n $, less than $ n+17 $, and coprime with the product $ n(n+17) $.