Let the sum be N

Write the last few terms of the sequence with the given sum as well:
 
We observe from the first few terms that only 
the multiples of 3 are negative and all other 
terms positive, so 96 and 99 are negative and 
97, 98,and 100 on the right end are positive.
 
1 + 2 - 3 + 4 + 5 - 6 + 7 + 8 - 9 + 10 + ... +95-96+97+98-99+100 =  N

Now we do the same with the sequence with the 
given sum. We observe from the first few that 
only numbers just before multiples of 3 are 
negative and all the others positive, so 95 
and 98 are negative and the others on the right 
end are positive.
 
  1 -  2 +  3 +  4 -  5 +  6 +  7 -  8  ... -95+96+97-98+99+100 = 1750

Now we write the terms of that sum in reverse order.  It will then, of
course, give the same sum of 1750.

100 + 99 - 98 + 97 + 96 - 95 + 96 - 95  ... + 6- 5+ 4+ 3- 2+  1 = 1750 

Now we add that term by term to the original equation:

  1+  2-  3+  4+  5-  6+  7+  8+ ... +95- 96+ 97+ 98- 99+100 =      N
100+ 99- 98+ 97+ 96- 95+ 94- 93+ ... + 6-  5+  4+  3-  2+  1 =   1750
---------------------------------------------------------------------
101+101-101+101+101-101+101-101+ ...+101-101+101+101-101+101 = N+1750   
    
There are 100 terms, so the sum on the left consists of 33 groups of 
three terms each (101+101-101) plus one 101 term on the right end, so 
the left side amounts to 33x(101+101-101)+101 or 33x101 + 101 = 34x101
= 3434

So the bottom equation becomes ---------->              3434 = N+1750
Solve 

3434 = N+1750

for N and we get N = 1684, which is choice (A).

Edwin