The sum of the squares of two consecutive positive odd numbers is 514.

Let the two consecutive positive odd numbers be x and (x + 2).

According to the given condition,

$x^{2}+(x+2)^{2}=514$

$\Rightarrow x^{2}+x^{2}+4 x+4=514$

$\Rightarrow 2 x^{2}+4 x-510=0$

$\Rightarrow x^{2}+2 x-255=0$

$\Rightarrow x^{2}+17 x-15 x-255=0$

$\Rightarrow x(x+17)-15(x+17)=0$

$\Rightarrow(x+17)(x-15)=0$

$\Rightarrow x+17=0$ or $x-15=0$

$\Rightarrow x=-17$ or $x=15$

∴ x = 15             (x is a positive odd number)

When x = 15,
x + 2 = 15 + 2 = 17

Hence, the required numbers are 15 and 17.