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:
$$\sum_{n}\langle x_i\rangle_{n}=\sum_n\int_0^1 x f_{in}(x)dx$$
The average momentum carried by any particular $n-$th Parton belonging to species $i$ is:
\begin{align}
\langle x_i\rangle_{n}&= \frac{\sum_n \langle x_i \rangle_n}{n}\\
&=\frac{\sum_{n}\int_0^1 xf_{in}(x)dx}{\sum_{n}\int_0^1 f_{in}(x)dx}\\
&= \int_0^1 xf_{in}(x)dx \frac{\sum_n 1}{\sum_{n}\int_0^1 f_{in}(x)dx}\\
&=\int_0^1 xf_{in}(x)dx 
\end{align}
The sum rule can be seen as:
$$\sum_{n}\underbrace{\int_{0}^{1}f_{in}(x)dx}_{=1} = \sum_{n}1= n$$
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.