Couldn't we just take $ f,g$ such that $ f(x)-g(x) = (x-1)^{100}$? Let $ (x-1)^{100}= \sum_{k=0}^{100}c_{k}x^{k}$. None of the $ c_{k}$'s is $ 0$, but $ \sum c_{k}= 0$. We construct the polynomials as follows: $ a_{100}= a; \ b_{100}= a_{100}-c_{100};$ $ a_{99}= b_{100}; \ b_{99}= a_{99}-c_{99};$ $ \vdots$ $ a_{0}= b_{1}; \ b_{0}= a_{0}-c_{0}$. We do have that $ a_{100}= b_{0}$, since $ \sum b_{k}= \sum \left( a_{k}-c_{k}\right) = \sum a_{k}$.