partition to min the max number of intersections
This is NP-hard, by reduction from 3-partition.

Given $\left[\hspace{.03 in}j_0,j_1,j_2,...,j_{(3\cdot k)-1}\hspace{-0.01 in}\right]$, $\:$ where $\; J = \operatorname{sum}\left(\left[\hspace{.03 in}j_0,j_1,j_2,...,j_{(3\cdot k)-1}\hspace{-0.01 in}\right]\right) \;$, $\;$ if there are

$k$ different stores, $\:J$ items, and $\:(3\cdot k)+(4\cdot J)\:$ customers,

with customer $i$ for $\:i\in [0,1,2,...,(3\cdot k)-1]\:$ interested in the obvious $j_i$ items

and four other customers interested in each one of the items,

then

the maximum can't be less than $\:\frac{4\cdot J}k+3\:$ and the partitions whose maximum is

$\frac{4\cdot J}k+3\:$ are exactly those corresponding to $3$-partitions of $\left[\hspace{.03 in}j_0,j_1,j_2,...,j_{(3\cdot k)-1}\hspace{-0.01 in}\right]$

.