a) Write $n = ab$, where $a$ and $b$ both are integers greater than $1$, and let $a_1 = a_2 = \dots = a_{n-1} = a$ and $a_n = 0$. The sum $a_1 + a_2 + \dots + a_n = a \cdot (n-1)$ is clearly not divisible by $n$. Consider any positive integer $k \le n$: If $k \le n-b$, then $a_k + a_{k+1} + \dots + a_{k+b-1}$ is divisible by $n$; otherwise, $a_k + a_{k+1} + \dots + a_{k+b}$ is divisible by $n$.

b) Suppose, if possible, that for each positive integer $k \le p$ there exists a positive integer $\ell \le p$ such that $a_k + a_{k+1} + \dots + a_{k+\ell-1}$ is divisible by $p$; to make a choice, let $f(k)$ be the least such $\ell$ (any choice will do). Notice that $f(k) \le p-1$, since $a_k + a_{k+1} + \dots + a_{k+p-1} = a_1 + a_2 + \dots + a_p$ is not divisible by $p$.

Define integers $0 = k_0 < k_1 < k_2 < \dots < k_p$ such that $k_{i+1} - k_i \le p-1$, $i = 0, 1, \dots, p-1$, and
$$a_{k_i+1} + a_{k_i+2} + \dots + a_{k_{i+1}} \equiv 0 \pmod{p}, \quad i = 0, 1, \dots, p-1.$$
The $k_i$ are recursively defined by $k_{i+1} = k_i + f(k_i - p\lfloor k_i / p \rfloor + 1)$, $i = 0, 1, \dots, p-1$.

Consider any integers $m$ and $n$ such that $0 \le m < n \le p$. Since $k_{i+1} - k_i \le p-1$ for all $i$, it follows that $k_n - k_m \le (n-m)(p-1) \le p(p-1)$. Moreover,
$$a_{k_m+1} + a_{k_m+2} + \dots + a_{k_n} = \sum_{i=m}^{n-1} (a_{k_i+1} + a_{k_i+2} + \dots + a_{k_{i+1}}) \equiv 0 \pmod{p}. \quad (*)$$
Amongst the $p+1$ integers $k_0, k_1, k_2, \dots, k_p$, there exist two, say $k_m < k_n$, such that $k_m \equiv k_n \pmod{p}$. Then
$$a_{k_m+1} + a_{k_m+2} + \dots + a_{k_n} = \frac{k_n - k_m}{p} \cdot (a_1 + a_2 + \dots + a_p) \not\equiv 0 \pmod{p},$$
since $(k_n - k_m)/p$ is a positive integer less than $p$, the sum $a_1 + a_2 + \dots + a_p$ is not divisible by $p$, and $p$ is prime. This contradicts $(*)$ and concludes the proof.