A dense NP complete language implies P=NP
This is a nice homework problem about Mahaney's theorem.

Note that the complement of a "dense" language is a sparse language. Moreover if a language is  $\mathsf{NP}$-complete its complement is $\mathsf{coNP}$-complete.

If there is a "dense" $\mathsf{NP}$-complete language, there is a sparse $\mathsf{coNP}$-complete language. 

Mahaney's theorem tells us that there is no sparse $\mathsf{NP}$-complete language unless $\mathsf{P}=\mathsf{NP}$. 

We can adopt the proof to show that there is no sparse $\mathsf{coNP}$-complete language unless $\mathsf{P}=\mathsf{coNP}$ which is equivalent to $\mathsf{P}=\mathsf{NP}$ (since $\mathsf{P}$ is closed under complements).

In summary, the answer is no unless $\mathsf{P}=\mathsf{NP}$. Note that if $\mathsf{P}=\mathsf{NP}$ then every nontrivial language is $\mathsf{NP}$-complete.

ps:
You may want to try the following and then use Mahaney's theorem: there is a sparse $\mathsf{NP}$-complete set iff there is a sparse $\mathsf{coNP}$-complete set.
However I doubt that a proof for this statement would be much easier than a proof for Mahaney's theorem.