$\mathbb{S} \subset \{0,1,2,...,2005\}$
Let $f : \{1,2,3, \cdots , 2005\} \rightarrow \mathbb{S}$ be such that $f(x) = \lfloor \frac{x^2}{2005} \rfloor$.
Claim : $f$ is onto function in $\{ 1,2,\cdots , 1002\}$ and one to one function in $\{ 1003,1004,\cdots,2005\}$.
Proof : Let $a = \lfloor \frac{x^2}{2005} \rfloor$ where $a \in \mathbb{S}$. From the property of greatest integer function, it can be concluded that $$a≤ \frac{x^2}{2005} < a+1 \iff 2005a ≤ x^2 < 2005(a+1) \iff \sqrt{2005a} ≤ x < \sqrt{2005(a+1)}$$
Now for $f$ to be onto, for every $a$ there exist $x$, i.e., there must be at least one integer between $\sqrt{2005(a+1)}$ and $\sqrt{2005a}$. This is possible when $$\sqrt{2005(a+1)} - \sqrt{2005a} ≥ 1 \implies 2005a ≤ 1002^2 \implies a ≤ \Big \lfloor \frac{1002^2}{2005} \Big \rfloor = 500 \Box $$
For $f$ to be one to one function, no value in $\mathbb{S}$ may be taken by $f(x)$ more than once. Therefore, $$\Big \lfloor \frac{x^2}{2005} \Big \rfloor \ne \Big \lfloor \frac{(x+1)^2}{2005} \Big \rfloor$$ $\frac{(x+1)^2}{2005} = \frac{x^2}{2005} + \frac{2x+1}{2005}$ Therefore when $\frac{2x+1}{2005}>1 \iff (2x+1) > 2005$ , or $x > 1002$, $f$ is one to one $\Box$.
Final Solution : $|\mathbb{S}| = 501 + (2005-1002) = \boxed{1504}$