Suppose they are a,b,c,and d, then their sums taken 3 at at time
are   

a + b + c =  72
a + b + d =  84
a + c + d =  96
b + c + d = 108

Let's line up all 4 or those equations and add them:

 a +  b +  c      =  72
 a +  b      +  d =  84
 a      +  c +  d =  96
      b +  c +  d = 108
-----------------------
3a + 3b + 3c + 3d = 360

Divide through by 3

    a + b + c + d = 120

That's the answer you were asked for.  

----------------------------

But we still don't know what those four integers are.
And they could have asked you that.

The given numbers 72, 84, 96, and 108 are consecutive 
multiples of 12, in fact they are:

6×12, 7×12, 8×12, 9×12

So if we can find 4 smaller integers which when added 3 at a time,
produce the sums: 6, 7, 8 and 9, then we can think of them in
"dozens" and multiply them by 12, and they will produce the 4 sums
above.

The smallest three integers 1,2, and 3 have sum 1+2+3 = 6
So 1+2+4 = 7, 1+3+4 = 8, and 2+3+4 = 9, so

1,2,3,4 when added 3 at a time, produce the sums: 6, 7, 8 and 9,
 
Multiplying them by 12, we have the integers 12, 24. 36, and 48.

Checking:

12+24+36 =  72
12+24+48 =  84
12+36+48 =  96
24+36+48 = 108 

And their sum is 12+24+36+48 =  120.

Edwin