Find the largest $k$ such that for every positive integer $n$ we can find at least $k$ numbers in the set $\{n+1, n+2, ... , n+16\}$ which are coprime with $n(n+17)$.
1. **Understanding the problem**: We need to find the largest \( k \) such that for every positive integer \( n \), there are at least \( k \) numbers in the set \(\{n+1, n+2, \ldots, n+16\}\) which are coprime with \( n(n+17) \).

2. **Analyzing the set**: Consider the set \(\{n+1, n+2, \ldots, n+16\}\). We need to check the gcd of each element in this set with \( n(n+17) \).

3. **Prime factorization**: Note that \( n(n+17) \) is a product of \( n \) and \( n+17 \). If \( d \) is a common divisor of \( n \) and \( n+k \) (where \( k \) is between 1 and 16), then \( d \) must divide \( k \). Since \( k \) ranges from 1 to 16, \( d \) can only be one of these values.

4. **Checking gcd conditions**: We need to ensure that for each \( n \), there are at least \( k \) numbers in the set \(\{n+1, n+2, \ldots, n+16\}\) such that \(\gcd(n+i, n(n+17)) = 1\).

5. **Special case analysis**: Consider \( n \) to be the product of the first 16 primes, i.e., \( n = 2 \cdot 3 \cdot 5 \cdot \ldots \cdot 47 \). In this case, \( n \) is divisible by all primes less than or equal to 47. Therefore, \( n+i \) for \( i \) from 1 to 16 will not be coprime with \( n \) if \( i \) is a multiple of any of these primes.

6. **Finding \( k \)**: We need to find the largest \( k \) such that there are at least \( k \) numbers in the set \(\{n+1, n+2, \ldots, n+16\}\) which are coprime with \( n(n+17) \). Since \( n \) is divisible by all primes up to 47, and \( n+17 \) is not divisible by any of these primes, we need to ensure that at least one number in the set is coprime with \( n(n+17) \).

7. **Conclusion**: Since \( n \) is divisible by all primes up to 47, and \( n+17 \) is not divisible by any of these primes, there will always be at least one number in the set \(\{n+1, n+2, \ldots, n+16\}\) which is coprime with \( n(n+17) \). Therefore, \( k \geq 1 \).

The final answer is \( \boxed{ k = 1 } \).