Solution:
Let us consider the set $\{1,2,\ldots,2017\}$ modulo $37$. Since $2017 = 37 \times 54 + 19$, the residues modulo $37$ that appear are $1,2,\ldots,37$ repeated $54$ times each, and $1,2,\ldots,19$ appearing one more time each.

Let us group the numbers by their residues modulo $37$.

For each pair of residues $r$ and $37 - r$ (with $1 \leq r \leq 18$), the sum $r + (37 - r) = 37$ is divisible by $37$. Thus, we cannot have both $r$ and $37 - r$ in $S$.

The residue $0$ (i.e., multiples of $37$) is its own pair, so at most one element with residue $0$ can be in $S$.

If $37$ is even, then $r = 37/2 = 18.5$, but since $37$ is odd, there is no such residue.

So, for each pair $\{r, 37 - r\}$ with $1 \leq r \leq 18$, we can take all numbers with residue $r$ or all with residue $37 - r$, but not both.

Let us count how many numbers there are for each residue:
- For $1 \leq r \leq 19$, there are $55$ numbers with residue $r$ (since $54$ full cycles plus one extra for $1 \leq r \leq 19$).
- For $20 \leq r \leq 37$, there are $54$ numbers with residue $r$.

Let us maximize the total size of $S$.

For each pair $\{r, 37 - r\}$ with $1 \leq r \leq 18$:
- $r$ has $55$ numbers if $1 \leq r \leq 18$.
- $37 - r$ has $54$ numbers if $19 \leq 37 - r \leq 36$ (i.e., $1 \leq r \leq 18$).

So, for each pair, we should pick the residue with more numbers, i.e., $r$ ($1 \leq r \leq 18$), which has $55$ numbers.

There are $18$ such pairs, so $18 \times 55 = 990$ numbers.

Now, consider residue $0$ (i.e., $37, 74, \ldots, 2017$):
- The numbers with residue $0$ are $37, 74, \ldots, 1998$.
- $2017 \div 37 = 54.5$, so $54$ multiples of $37$ up to $1998$.
- $37 \times 55 = 2035 > 2017$, so only $54$ such numbers.
- So, at most one can be chosen.

Residue $19$ is left unpaired (since $37 - 19 = 18$ is already paired with $19$). Residue $19$ has $55$ numbers.

Thus, the maximum is:
$$18 \times 55 + 55 + 1 = 990 + 55 + 1 = 1046.$$

Therefore, the answer is $\boxed{1046}$.