4. Find all natural numbers $n$ for which all numbers $n+1, n+3, n+7$, $n+9, n+13$ and $n+15$ are prime.
4. The only such number is $n=4$ (the observed numbers are then $5,7,11,13,17$ and 19).

We claim that one of the numbers $n+1, n+3, n+7, n+9, n+13$ and $n+15$ is divisible by 5. Indeed, depending on whether $n$ gives a remainder of 0, 1, 2, 3, or 4 when divided by 5, the number divisible by 5 will be $n+15, n+9, n+3, n+7$ or $n+1$, respectively. Therefore, since all the observed numbers are prime by the condition of the problem, and one of them is divisible by 5, it follows that one of the numbers must be exactly equal to 5. This is clearly possible only for the numbers $n+1$ and $n+3$. In the first case, we get $n=4$, which is the solution already mentioned. In the second case, we get $n=2$, but then we have $n+7=9$, which is not a prime number, so this is not a solution.