The problem with this approach is that we are treating $f_i(x)$ purely as probability distribution function. As such, a different normalisation condition would have been imposed over it. Refering to "Fundamentals of Particle Physics" by Pascal Paganini
They represent the probability densities (strictly speaking, they rather represent the number densities as they are normalised to the number of partons) to find a parton of type $i$ carrying a momentum fraction $x$ of the proton. In other words, $f_i(x) dx$ is the number of partons of type $i$ within the proton carrying a momentum fraction between $x$ and $x + dx$.
The correct approach to understanding PDF would be following. The average momentum carried by $n$ partons (number of parton - number of anti-parton = $n$) of species $i$ is given by: $$\sum_{n}\langle x\rangle_{n}=\sum_n\int_0^1 x f_{n}(x)dx$$ The average momentum carried by any particular Parton belonging to species $i$ is: $$\langle x\rangle_{n}= \frac{\sum_n \langle x \rangle_n}{n}=\frac{\sum_{n}\int_0^1 xf_{n}(x)dx}{\sum_{n}\int_0^1 f_{n}(x)dx} $$ The sum rule can be seen as: $$\sum_{n}\underbrace{\int_{0}^{1}f_{n}(x)dx}_{=1} = \sum_{n}1= n$$ Instead of having to rewrite the $`\sum_n'$ over and over again, we redefine the PDF by suppressing the index $n$ and the associated sum. Under this redefinition, we use the sum rule as normalisation condition over the PDFs. Returning to the problem in hand, \begin{align} \sum_i\sum_n \langle x_i\rangle_n &=\sum_{i}\sum_{n}\frac{\sum_{n}\int_0^1xf_{in}(x)dx}{\sum_{n}\int_0^1 f_{in}(x)dx}\\ &=\sum_{i}\underbrace{\sum_{n}\int_0^1xf_{in}(x)dx}_{\int_0^1 xf_{i}(x)dx}\frac{\sum_{n}1}{\sum_{n}\int_0^1 f_{in}(x)dx}\\ &= \sum_i \int_0^1 xf_{i}(x)dx=1 \end{align} We see that there is no contradiction here.