A subset of 2017 elements

Let $A_k$ denote the set of numbers from $1$ to $2017$ whose remainder is $k$ upon division by $37$ .

Since $2017$ is not a multiple of $37$ , these sets are not all the same size; $A_k$ has $55$ elements when $1 \le k \le 19$ , and $54$ elements otherwise.

$S$ can contain at most one element of $A_0$ .

$S$ can have as many elements from $A_1$ as we want, as long as it doesn't have any element from $A_{36}$ , and so on; we can choose to have elements from $A_k$ or $A_{37-k}$ , but not both. Clearly we want to pick elements from the bigger sets when we can; these are $A_1,\ldots,A_{18}$ (we could pick $A_{19}$ instead of $A_{18}$ , but it doesn't change the total).

The maximum number of elements is therefore $1+18\times 55=\boxed{991}$ .

I've seen this problem or variants of it before - does anyone know if it has a name?