{upper is entered from user}
  initialize sum to 0;
  for index := 1 to upper do
    if index is an odd number then
      add it to sum;
  report results of sum to user;

That's all you have to do.  The most difficult part here (which shouldn't even
be that difficult) is figuring out how to test for odd/even using "mod 2".
Bonus hint:  In case you didn't know, the "mod" operator returns the remainder
of division.  When you learned elementary math, what can you determine by
dividing a number by 2?

var
  cointer, sum, number : integer;

But there is an alternative solution which is much faster when N is large.
Suppose there is a second problem which ask the sum of 1 to 100.
Instead of using a loop to sum that one by one, we can simply use
symmetry
   1  2  3    ........    98  99 100
There are 50 pairs of {[100 + 1], [99 + 2], [98 + 3], ...}.
So the answer will be 101*50=5050.

Return back to the original problem (i.e. sum odd from 1 to N).
Here is an alternative solution using similar symmetry trick, as below.

Function sum_odd_by_symmetry ( N : integer) : integer;
var
  tmp, M, sum : integer;
begin
  if odd(N) then
    begin
      tmp := (N + 1) div 2;   { N + 1 equals sum of such a pair}
                       { Divide it by 2 to count only half of odd & even}
      sum := tmp * tmp; { tmp also equal the number of such pairs}
    end
  else
    begin
      M := N - 1;      {if N is even, count it down by 1 to get the odd}
      tmp := N div 2;  { N  equals sum of such a pair}
                       { Divide it by 2 to count only half of odd & even}
      sum := tmp * tmp; { tmp also equal the number of such pairs}
    end;
  sum_odd_by_symmetry := sum;
end;