Answer
$$x=8,\ y=8$$
Work Step by Step
Let $x,\ y $ two positive numbers such that
$$x+y=16$$
Then $y=16-x $
Since the sum of squares given by
$$S=x^2+y^2=x^2+(16-x)^2=2x^2-32x+256$$
The goal is to minimize $S$ , since
$$ S'(x)=4x-32$$
Then $S'(x)=0$ for $x=8$ , since $S'(x)<0$ for $x<8$ , $S'(x)>0$ for $x>8$, hence $S(x)$ has minimum at $x=8$ and $y=16-x=8$