Q:

Inconsistency with electrostatic energy formulas

The energy of point charge configuration can be written as:
$$W = \frac{1}{2}\sum_{i=1}^{n}q_{i}V(r_{i}) \, ,$$
which can take both positive and negative values. 
However, when we integrate the equation to get the energy of a continuous charge dustribution:
$$W = \frac{1}{2}\int\rho Vd\tau \Rightarrow W = \frac{\epsilon_{0}}{2}\left [ \int E^{2} d\tau + \oint VE\cdot da\right ] \, .$$
Take the volume to integrate to be all space, then the second term vanishes:
$$W = \frac{\epsilon_{0}}{2}\int E^{2}d\tau \, .$$
This formula can take only positive values.  
So there is a discrepancy between the two formulas.  What caused the discrepancy? 
According to Griffith's Introduction to Electrodynamics, it says in the former equation $V(r_{i})$ represent the potential due to all charges but $q_{i}$, whereas later $V(r_{i})$ is the full potential.  But why would the original charge has any potential when there is no other charge already present? 

A:

The difference is the zero point.  When summing over charges, the reference is a state in which this charges are infinitely separated.  Those are still distinct, localized charges, just separated from each other.  
When integrating $E^2$ over all space, the reference state has all charge separated.  Even the individual charges from the first method are broken up, so that $E=0$ everywhere.  
The second method has its reference at an "absolute zero", so to speak.  The first method has its reference with a lot of positive energy needed to gather the individual charges.  That's why the first method can have negative values. 