Separated 3Sum versus 3Sum problem
You can reduce the SEPARATED k-SUM problem to a (k+1)-SUM problem in the following way:

Given $X_1,...,X_n$, let $b$ such that $2^b &gt; \max( abs(X_1) \cup ... \cup \;abs(X_n))$, where $abs(X_i) = \{ |x| \mid x \in X_i \}$, and let $c$ such that $2^c &gt; k 2^b$ 

For  $i = 1,...,k$ build $Y_i = \{ 2^{c+ki} + 2^{b} + x_j \mid x_j \in X_i \}$;
let $u = - \sum_{i=1}^k 2^{c+ki} -k2^b$

Build an equivalent (k+1)-SUM problem picking $X =  \{ u \} \cup Y_1 \cup ... \cup Y_k$

Note that every $Y_i$ contains only positive elements.

Informally we require a $k+1$ elements subset $C$ from $X$ whose elements sum to zero.
We cannot build a solution only with the negative element $u$; so we must include in $C$ at least one element $y$ from one of the $Y_i$; but the $2^{c+ki}$ "component" of $y$ can be balanced only if $u$ is also included; but including $u$ implies that exactly one of the elements of each $Y_i$ is included (the  $- \sum_{i=1}^k 2^{c+ki}$ component of $u$ allows to keep the separation between the elements of the original $X_i$). But we also have $\sum_{j=1}^k (2^b + x_j) = k2^b$ and this implies that the original $x_j$s (coming from distinct $X_i$s as seen above) sum to zero.