Is this partition problem strongly NP-complete? Some computational problems have variants that appear to be harder. For instance, Graph Automorphism (GA) problem has quasi-polynomial time algorithm ( by Babai's Graph Isomorphism result) while the fixed-point free GA problem is NP-complete. 
Partition problem is weakly NP-complete problem since it has pseudo-polynomial time algorithm. I am interested in variants that are strongly NP-complete.
Here is a variant of partition problem:
Restricted partition problem
Input: Set $S$ of $2N$ integers, and a collection of pairs $P$ from $S$ 
Query: Is there a partition of $S$ into two equal cardinality parts $A$ and $S-A$ such that both parts have the same sum and no pair in $P$ has both elements in one side of the partition?

Is this variant of partition problem NP-complete in the strong sense? 

 A: It seems, that there exists a pseudo-polynomial time algorithm for your problem.
Denote by $N$ the sum of $S$. Add $N \cdot |S|$ to each element of $S$. This allows us to not carry about the cardinality restriction. 
Consider a square boolean table $T$ of size $N^2 \cdot |S|$  with the following sense: 
$T(X,Y)=1$ if there exist two  disjoint subsets $S_X$ and $S_Y$ of $S$ such that no pair in $P$ has both elements in $S_X$ or in $S_Y$ and $|P \cap (S_X \cup S_Y)| \not= 1 $.
Now use standard dynamic programming. 
First fill all elements of $T$ by $0$.
Fill $T(0,0):=1$.
Assume we have filled a part of the table.
Let $a$ be an element of $S$ that we have not considered yet. If $a$ is not element of a pair  then 
fill $T(X + a, Y) = T(X, Y+a)=1$ for all $X$ and $Y$ such that $T(X, Y)=1$.
If $(a,b)$ is a pair then $T(X + a, Y+b) = T(X+b, Y+a)=1$.
After  filling all table the answer is the element  $T(\frac{N\cdot|S|^2 + N}{2},\frac{N\cdot|S|^2 + N}{2})$ 
since the all sum now is $N\cdot|S|^2 + N$.
