Let $p$ be a prime and let $a_1, \ldots, a_{p-1}$ be natural numbers such that no $a_i$ is divisible by $p$. For a subset $I$ of $\{1, \ldots, p-1\}$, write $a_I$ to denote $\sum_{i\in I}a_i$, where $a_I$ is to be interpreted as $0$ when $I$ is the emptyset.
Want to show that: For any $0\leq x\leq p-1$, there is $I\subseteq \{1, \ldots, p-1\}$ such that $a_I\equiv x\pmod{p}$.
I tried the following approach. Suppose we just want to show that there is $I\subseteq \{1, \ldots, p-1\}$ such that $a_I\equiv 1\pmod{p}$. So the product $$\prod_{I\subseteq \{1, \ldots, p-1\}} (a_I-1)$$ must be shown to be equal to zero. Since this needs to be proved whenever none of the $a_i$'s are divisible by $p$, we need to show that the polynomial in $\mathbb F_p[x_1, \ldots, x_{p-1}]$ defined by $$\prod_{I\subseteq \{1, \ldots, p-1\}, I\neq \emptyset} (x_I-1)$$ takes the value $0$ whenever none of the $x_i$'s are $0$. This is equivalent to showing that the polynomial $$\prod_{i=1}^{p-1}x_i\prod_{I\subseteq \{1, \ldots, p-1\}, I\neq \emptyset} (x_I-1)$$ always takes the value $0$, which is same as showing that the above polynomial lies in the ideal $\langle x_1^p-x_1\rangle + \cdots+\langle x_{p-1}^p-x_{p-1}\rangle$.
I was able to carry this out only in the examples when $p=3, 5$, but all I did was explicitly calculate the polynomial and replace $x_i^p$ by $x_i$.
I am unable to do this for general $p$.
Considering $a_1,\dotsc, a_{p-1}$ as elements of $\mathbb Z_p$, for each $0\le k\le p-1$ let $S_k$ denote the set of all those $z\in\mathbb Z_p$ representable as a linear combination $z=\varepsilon_1a_1+\dotsb+\varepsilon_ka_k$ with the coefficients $\varepsilon_1,\dotsc,\varepsilon_k\in\{0,1\}$. Thus, for instance, $S_0=\{0\}$, $S_1=\{0,a_1\}$, and we want to show that $S_{p-1}=\mathbb Z_p$. To this end, observe that $S_{k+1}=S_k\cup(S_k+a_{k+1})$, and that $S_k+a_{k+1}\ne S_k$, unless $S_k=\mathbb Z_p$. (This is not completely trivial, but not difficult either.) Thus, we have either $S_k=\mathbb Z_p$, or $|S_{k+1}|\ge|S_k|+1$. Hence, arguing inductively, either $|S_{k+1}|\ge k+1$, or $S_{k+1}=S_k=\mathbb Z_p$. In particular, for $k=p-1$ we get $S_k=\mathbb Z_p$, as wanted.