30. For all $n$ such $N$ exists. For a given $n$ choose $N=(n+1)!^{2}+1$. Then $1+j$ is a proper factor of $N+j$ for $1 \leq j \leq n$. So if $N+j=p^{m}$ is a power of a prime $p$, then $1+j=p^{r}$ for some integer $r, 1 \leq r<m$. But then $p^{r+1}$ divides both $(n+1)!^{2}=N-1$ and $p^{m}=N+j$, implying that $p^{r+1} \mid 1+j$, which is impossible. Thus none of $N+1, N+2, \ldots, N+n$ is a power of a prime. Second solution. Let $p_{1}, p_{2}, \ldots, p_{2 n}$ be distinct primes. By the Chinese remainder theorem, there exists a natural number $N$ such that $p_{1} p_{2} \mid N+1, p_{3} p_{4} \mid N+2, \ldots, p_{2 n-1} p_{2 n} \mid N+n$, and then obviously none of the numbers $N+1, \ldots, N+n$ can be a power of a prime.