SOLUTION: If the sum of the first 100 integers: 1+2+3...+99+100=5050, then the sum of the 50 odd integers: 1+3+5...+99=?

Let the sum of the first fifty odd numbers be N

  1  +   3   +   5   + ··· +  95  +  97  +  99  =  N

Write the above equation down, then right under them,
write the same numbers added in the reverse order. 
Like this:

  1  +   3   +   5   + ··· +  95  +  97  +  99  =  N
 99  +  97   +  95   + ··· +   5  +   3  +   1  =  N
----------------------------------------------------
       

Now add those two equations term by term, like this:

  1  +   3   +   5   + ··· +  95  +  97  +  99  =  N
 99  +  97   +  95   + ··· +   5  +   3  +   1  =  N
----------------------------------------------------
100  + 100   + 100   + ··· + 100  + 100  + 100  = 2N

There are fifty terms, all 100's in the bottom equation,
so obviously that's 50x100 or 5000, so the above becomes:

                                           5000 = 2N

Divide both sides by 2                     2500 = N

So the sum is 2500.

(You didn't need the sum of the first 100 positive integers.)

Edwin