Can a charge-pump be 100% efficient, given ideal components?
It is all about dualism. With ideal components, you can make an ideal SMPS type voltage converter (= using an inductor to do the work). You can't make an ideal voltage converter using switched (flying) capacitors. That is not the universe being unfair to capacitors: you can make an ideal current converter using switched capacitors, which is not possible using inductors.
The problem with capacitors and a voltage source is like this: take a voltage source \$V\$ with a certain source impedance (= series resistor) \$R\$. Connect a capacitor \$C\$ to it and load it for an infinite time (any finite time will do too). The loading current will be
$$ I = \frac{V}{R}\exp\left(\frac{-t}{RC}\right)$$
and so the power dissipated over the resistor will be
$$ P = I^2 R = \frac{V^2}{R}\exp\left(\frac{-2t}{RC}\right)$$
Consequently, the total energy cost shall be
$$ E = \int_0^\infty P \,dt =  \frac{V^2}{R} \int_0^\infty \exp\left(\frac{-2t}{RC}\right) \,dt = \frac{CV^2}{2} $$
which, as you can see, is both always positive and completely independent of \$R\$. Therefore, there is always a cost, and even with an ideal capacitor! Intuitively, this is because a smaller resistor causes a higher initial loading current, and hence a higher RI2 loss.
Management summary:
You can't connect an ideal voltage source to a capacitor, because that would result in an infinite current which is impossible in itself and would cause an infinite magnetic field which would destroy the universe (just kidding, remember this is the management summary). But you can approach this ideal as closely as you like, and the result will still be the same: a fixed amount of energy is lost while charging the capacitor. Hence: sorry boss, no ideal flying capacitor voltage converter.