First prove that $1$ and $-1$ are not expressible as the sum of at least three consecutive terms of an arithmetic sequence of integers. Let $a_1$, $a_2$, $\ldots$, $a_k$ be $k$ consecutive terms of an arithmetic sequence, where $k \ge 3$. They sum up to $s = \frac{a_1 + a_k}{2} \cdot k$. If $k$ is odd, then $s$ is divisible by $k$. If $k$ is even, then $s$ is divisible by $\frac{k}{2} > 1$. In both cases, $s$ differs from $1$ and $-1$.

Now prove that every integer $s$ other than $1$ or $-1$ is expressible as the sum of at least three consecutive terms of an arithmetic sequence of integers. If $s = 0$, then $s = -1 + 0 + 1$. If $s$ is different from zero and is even, i.e., $s = 2t$, where $t \neq 0$, then $-t, 0, t, 2t$ sum up to $2t = s$. If $s$ is odd, i.e., $s = 2t + 1$, then $-t + 1, \dots, 0, 1, \dots, t-1, t, t+1$ are consecutive terms of an arithmetic sequence; they sum up to $t + (t+1) = 2t + 1 = s$, since the terms $-t+1$ through $t-1$ mutually cancel.

Let $a_1$ be the first of the consecutive terms and $d$ be the common difference of consecutive terms. The sum of $n$ consecutive terms is $s = \frac{2a_1 + d(n-1)}{2} \cdot n$. Thus $2s = (2a_1 + d(n-1))n$. If $s = 1$ or $s = -1$, then this equality cannot hold because $n \ge 3$ divides neither $2$ nor $-2$. If $s = 0$, then choose the portion of the arithmetic progression to be $-1$, $0$, $1$. If $s$ differs from these numbers, then let $n = 2|s|$, $d$ be an arbitrary odd number, and $a_1 = \frac{1 - (n-1)d}{2}$ if $s > 0$, and $a_1 = \frac{-1 - (n-1)d}{2}$ if $s < 0$.