Solution:

Suppose that $2005 = a + b$, where $a$ is formidable and $b$ is successful. Then $a$ must be a sum of some of the powers of $4$ less than $2005$, namely $1, 4, 16, 64, 256, 1024$. Similarly, $b$ must be a sum of some of the numbers $1, 6, 36, 216, 1296$. So, $a + b$ must be a sum of some distinct entries from the list
$$1, 1, 4, 6, 16, 36, 64, 216, 256, 1024, 1296.$$
If we use both $1024$ and $1296$, we get at least $1024 + 1296 = 2320$ which is too big. But if we omit one of them, the most we can get is
$$1 + 1 + 4 + 6 + 16 + 36 + 64 + 216 + 256 + 1296 = 1896$$
which is too small. So there is no way to achieve a sum of $2005$.