jmc

Setting $y=-x,$ we get $$2f(x^2)=f(0{)}^2+2x^2$$for all $x.$ Setting $x=0$ in this equation, we get $2f(0)=f(0{)}^2,$ so $f(0)=0$ or $f(0)=2.$

Suppose $f(0)=2.$ Then $$2f(x^2)=4+2x^2,$$so $f(x^2)=x^2+2$ for all $x.$ In other words, $f(a)=a+2$ for all $a\ge 0.$

Setting $x=y=1$ in $f(x^2)+f(y^2)=f(x+y{)}^2-2xy,$ we get $$1^2+2+1^2+2=(2+2{)}^2-2\cdot 1\cdot 1,$$which simplifies to $6=14,$ contradiction.

Otherwise, $f(0)=0.$ Then $2f(x^2)=2x^2,$ so $f(x^2)=x^2$ for all $x.$ In other words, $f(a)=a$ for all $a\ge 0.$

Setting $y=0$ in $f(x^2)+f(y^2)=f(x+y{)}^2-2xy,$ we get $$f(x^2)=f(x{)}^2.$$But $f(x^2)=x^2,$ so $f(x{)}^2=x^2.$ Hence, $f(x)=\pm x$ for all $x.$

Then the given functional equation becomes $$x^2+y^2=f(x+y{)}^2-2xy,$$or $$f(x+y{)}^2=x^2+2xy+y^2=(x+y{)}^2.$$We have already derived this, so as far as the given functional equation is concerned, the function $f(x)$ only has meet the following two requirements: (1) $f(x)=x$ for all $x\ge 0,$ and $f(x)=\pm x$ for all $x<0.$

Then we can write $$\begin{array}{rl}S& =f(0)+(f(1)+f(-1))+(f(2)+f(-2))+(f(3)+f(-3))+\cdots +(f(2019)+f(-2019))\\ & =2(c_1+2c_2+3c_3+\cdots +2019c_{2019}),\end{array}$$where $c_i\in \{0,1\}.$ We can check that $c_1+2c_2+3c_3+\cdots +2019c_{2019}$ can take on any value from 0 to $\frac{2019\cdot 2020}{2}=2039190,$ giving us $\boxed{2039191}$ possible values of $S.$