The meaning of that $\ \vert\ $ in this context is divides (as in evenly divides, or is a divisor of), and that's a standard usage of this sign. The quote should be read as:
let $p=311$ and $r=31$, which divides $(p-1)$. Let $g=169$…
In other words: $r$ is a divisor of $p-1$. Or, exists integer $q$ with $r\times q=p-1$. Or, $((p-1)\bmod r)=0$, also writable as $p-1\bmod r=0$ or $p-1\equiv0\pmod r$ or $p\equiv1\pmod r$. In many common programing languages, (p-1)%r == 0. That's because $31$ (evenly) divides $311-1$, since $31\times10=310$.
That was correctly guessed by Aman Grewal in comment, but as noted, proximity with the assignment makes the notation confusing. Elision of the implied which is something I would try to avoid.
The end of the sentence says «Let $g=169$, which has order $r$». Does that mean that $g$ is really 169%31?
No. The term order is used in its meaning in group theory. In this context, it means that when we repeatedly multiply $1$ by $g$, reducing modulo $p$ after each multiplication, we'll first get back to $1$ after performing $r$ multiplications. That's related to $r\,\vert\,(p-1)$, because the order of any element in a finite group is a divisor of the order of the group, that is the number of elements in the group. Here the group is the multiplicative group modulo $p$, noted $\mathbb Z_p^*$ or $(\mathbb Z/p\mathbb Z)^\times$, which has $p-1$ elements since $p$ is prime. The powers of $g$ form a subgroup of order $r$, called a Schnorr group.
