nc was intended to count the different sums. For the time being it is not important. Concerning your question and using Rohit's method

sum = 0;
sums = Table[
    sum = sum + a[i][j]*If[j == 2, 0, 1], {i, 1, 3}, {j, 1, 3}] // 
   Flatten;
sums // Column

But that is exactly what is given by the code given above.

Like this?

nc = 0;
sum = 0;
For[i = 1, i < 4, i = i + 1,
 For[j = 1, j < 4, j = j + 1,
  sum = sum + a[i, j];
  Print[sum]
  ]
 ]

It gives 9 lines of output, and instead of Print[ sum ] you may store it somewhere.

Another way

sum = 0;
sums = Table[sum = sum + a[i, j], {i, 1, 3}, {j, 1, 3}] // Flatten

sums // Column

Well, I do not understand what you want now. My code produces exactly that what you asked for in your initial post. What about a little "Thank you"?

By the way, a[ i ][ j ] could be interpreted as mappings of R2 -> R1. A solution of an equation like

f[ ...., a[i][j],....] ==0

cannot have the form

{n,m}

unless f [ .... ] is an element of R2

But if you want to add a previous result to the next one you coul do it like this

ff[j_] :=  a[1 + Floor[(j - 1)/3]][ If[Mod[j, 3] == 0, 3, 1]] (If[Mod[j + 1, 3] != 0, 1, 0])

out = Table[0, {9}];
out[[1]] = xx;
out[[2]] = xx;
out

Do[
 out[[j]] = out[[j - 1]] + f[j],
 {j, 3, 9}]
out

out /. f -> ff /. xx -> a[1][1];
% // TableForm

In this case

a[i][j]==0 for j=2.

if it is not zero output will be

a[1][1] then a[1][1]+a[1][2] then a[1][1]+a[1][2]+a[1][3] then a[1][1]+a[1][2]+a[1][3]+a[2][1] and so on. 

I need this. After each output I need to solve the system of equation like

{a[1][1]*1,a[1][1]*1}=={2,2} then

{a[1][1]*1 + a[1][2]*1,a[1][1]*1 + a[1][2]*2}=={2,3} then

{a[1][1]*1 + a[1][2]*1 +a[1][3]*1,a[1][1]*1 + a[1][2]*2 +a[1][3]*3}=={3,4}

and so on. Range of i and j may be upto 20.