Balancing using oxidation numbers
It is really easy to solve by the half reaction method, here is my answer:


Half reaction method

Global eq: KCl + MnO2 + H2SO4 -> K2SO4 + MnSO4 + Cl2 + H2O

Cathode: 2Cl- &nbsp; --> Cl2 + 2e-

Anode: MnO2 + 4H+ + 2e- --> Mn+2 + 2H2O

So the final balanced eq is this one:

2KCl (ac)  + MnO2 (s) + 2H2SO4 (ac) --> K2SO4 (ac) + MnSO4 (ac) + Cl2 (g)  + 2H2O



Variations method

Without half-reaction method will be like this:
You have one Mn+4 reduced to Mn+2
That is 2e x 1 atom (ion indeed) so this is a net variation of 2
On the other hand you have Cl- wich oxidizes to form a Cl2 molecule so this is
1e x 2 atoms wich also is a net variation of 2
Now you use these numbers as the coefficient of the products that contain these atoms and you'll have  

KCl (ac)  + MnO2 (s) + H2SO4 (ac) --> K2SO4 (ac) + 2MnSO4 (ac) + 2Cl2 (g)  + H2O

and considering this as valid just left to adjust the rest of the elements
and finally you'll have:
4KCl (ac)  + 2MnO2 (s) + 4H2SO4 (ac) --> 2K2SO4 (ac) + 2MnSO4 (ac) + 2Cl2 (g)  + 4H2O
Wich is also a valid adjust

I always recommend using half-reaction method or undetermined coefficient method instead. Good luck!!