Hardness proof of EVEN-ODD PARTITION
Let $b_i = a_{2i} - a_{2i-1}$. The problem is then finding $\sum \varepsilon_i b_i = 0$. Here $\varepsilon_i = 1$ if $a_{2i} \in A'$ and $-1$ otherwise. Let $B' = \{b_i \mid \varepsilon_i = 1\}$. Note that $\sum_{b \in B'} b = \sum _{b \in B \setminus B'} b$. Thus solving PARTITION of $B$ would solve EVEN-ODD PARTITION of $A$.

Let $b_{2i} = a_i$ and $b_{2i-1} = 0$. Note that if $b_{2i} \in B' \leftrightarrow a_i \in A'$ and $\sum_{B'} = \sum_{B \setminus B'}$ then $\sum_{A'} = \sum_{A \setminus A'}$. Thus solving EVEN-ODD PARTITION on $B$ would solve PARTITION on $A$.