Is the $k$P$k$N-3SAT problem NP-complete?
1) 2P2N-3SAT is NP-complete

(Edit to take into account all comments below)

Take a 2P1N-SAT instance, check the all-1 vector is not a model, then add the clause with all variables negated to the formula to transform it into a 2P2N-SAT instance. 

For each 1-clause $(a)$ in the new 2P2N formula, replace it by the three following 3-clauses : $(a \lor x \lor x)(\lnot x \lor y \lor y)(\lnot x \lor \lnot y \lor \lnot y)$ where $x,y$ are fresh 2P2N variables (i.e. appearing twice as positive literals and twice as negative literals).
For each 2-clause $(a \lor b)$ in the formula, replace it by the six following 3-clauses : $(a \lor b \lor x)(\lnot x \lor y \lor y)(\lnot x \lor \lnot y \lor \lnot y)(x \lor z \lor \lnot z)(z \lor t \lor  \lnot t)(\lnot z \lor t \lor \lnot t)$ where $x,y,z,t$ are fresh 2P2N variables.
Keep all 3-clauses in the formula.
For each $k$-clause ($k&gt;3$) in the formula, transform it into a set of 3-clauses by the usual way, which does not change the number of occurences of positive (negative) literals of the taken $k$-clause, but add only extra variables which appear once as a positive literal and once as a negative literal. For each such new variable $x$, add the two following clauses to the 3-CNF formula :  $(x\lor y \lor \lnot y)$ and  $(\lnot x \lor \ y\lor  \lnot y)$ where $y$ is a fresh 2P2N variable. 

All these transformations can obviously be done in polynomial time and transform the 2P1N-SAT instance into a 2P2N-3SAT instance. The NP-complete 2P1N-SAT problem is then reducible to the 2P2N-3SAT problem.

2) The extension to $k$P$k$N-3SAT can be done in the same spirit (a bit longer) to prove it is NP-complete as well.