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Just let each thread pull pairs of (a,b) and compute the partial result until all pairs are consumed.

Main thread:

output = 0
iter = iterator_over(A,B)

// start threads and wait until done

answer = output / size(A) / size(B)
return answer

Each thread:

res = 0
while true:
   synchronized:
       if iter.hasNext():
           a,b = it.next()
       else:
           break

   res += f(a, b, n)

synchronized:
    output += res

For optimal performance, the amount of threads should be the same as the amount of virtual cores.