That is for sum of two cores. It might be proved by indicating the number of the used core in ParallelEvaluate[]:
In[17]:= ClearAll[x]
x[] := RandomInteger[10, {100, 1000}]
x[] // ByteCount
ParallelEvaluate[MaxMemoryUsed[total = Sum[x[], {i, 1000}]], 1]
Out[19]= 800152
Out[20]= 800977040
For ParallelSum[]:
In[25]:= x[] // ByteCount
MaxMemoryUsed[total = ParallelSum[x[], {i, 1000}]]
Out[25]= 800152
Out[26]= 25133408
Sure, Do[] is better:
In[17]:= total = 0;
MaxMemoryUsed@Do[total += x[], 1000]
Out[18]= 2400640
or Nest[]:
In[27]:= MaxMemoryUsed@Nest[Plus[x[] + #] &, 0, 1000]
Out[27]= 2400880
or While[]:
In[50]:= total = x[]; i = 0;
MaxMemoryUsed@While[++i < 1000, total += x[]]
Out[51]= 1600384