The number of octahedral and tetrahedral sites in a cubical close packed array of N spheres respectively is

Number of octahedral sites = number of spheres in packing 

Number of tetrahedral voids $=2\times$ number of spheres in packing 

Number of tetrahedral voids $=2\times$octahedral voids 

So, the answer is N and 2N.

NOTE : For  fcc arrangement

 THV'S are located along body diagonal and on each body diagonal 2 THV'S are present.

Therefore Number of THV'S per unit cell =$4\times 2\times 1=8$

Hence in fcc , Number of THV'S is twice to the number of particles.

$\Rightarrow$OH'S are located at edge centre and body centre .

 Number of octahedral  per unit cell =$12\times \frac{1}{4} + 1\times 1 = 4$

Hence in fcc , Number of OHV'S is equal  to the number of particles.