Exact complexity of a problem in $\cap_{m \geq 2}\mathsf{AC}^0[m]$ Let $x_i \in \{-1,0,+1\}$ for $i \in \{1,\ldots,n\}$, with the promise that $x = \sum_{i=1}^n{x_i} \in \{0,1\}$ (where the sum is over $\mathbb{Z}$). Then what is the complexity of determining if $x = 1$?

Notice that trivially the problem lies in $\cap_{m \geq 2}{\mathsf{AC}^0[m]}$ because $x \equiv 1\bmod{m}$ iff $x = 1$. Question is: does the problem lie in
$\mathsf{AC}^0$? If so, what is the circuit witnessing this? If not, how does one prove this?
I do not think this is in AC0 and I can show a lower bound for the related promise problem of distinguishing between $\sum x_i = 0$ and $\sum x_i = 2$, when $x \in \{-1, 1\}^n$. Similar Fourier techniques should apply to your problem, but I have not verified that. Or maybe there is a simple reduction. 

Suppose there is a size $s$ depth $d$ circuit that computes a function $f: \{-1, 1\}^n \rightarrow \{0, 1\}$ such that  $f(x) = \sum_i x_i$ whenever $\sum_i x_i \in \{0,2\}$. Because for a random $x$, the probability that $\sum_i x_i = 0$ is $2^{-n} {n \choose n/2} \approx n^{-1/2}$, and for each such $x$ there are $n/2$ coordinates that change the value of $f$, the total influence of $f$ is $\Omega(n^{1/2})$, which is roughly the same as majority (because you included most of majority's sensitive inputs). By a theorem of Hastad (see Colorraly 2.5 in Ryan O'Donnel's notes), this implies that 

$$
s \geq 2^{\Omega(n^{1/(2d-2)})}.
$$