Solution:

Consider the following addition:
$$\begin{array}{r} 111\cdots 100\cdots 01 \\ + \\ 11\cdots 11 \end{array}$$
By making the blocks of 1's and 0's appropriately long, we can ensure that the addends respectively contain 2004 and 2005 1's. (To be precise, we get $a = 2^{4008} - 2^{2005} + 1$ and $b = 2^{2005} - 1$.) Then the sum has only one 1. And clearly it is not possible to get any less than one 1.