I am searching for a result in the literature that I am sure must be known, but I just fail to find it.


Let us starts with a simple example:
Let $A, B\subset \mathbb{Z}$ be a finite sets of integers such that $|A|=|B|=2m+1$ and denote by $I$ the set $\{-m,...,m\}$. I want to show that for all $k,l \geq 0$ we have 
$$\sum\_{|i|\leq k, |j|\leq l}|(A+i)\cap(B+j)|\leq \sum\_{|i|\leq k, |j|\leq l}|(I+i)\cap(I+j)|.$$


And, more generally, for any finite collection of sets $A\_1,\ldots,A\_t$ of the same odd cardinality $2m+1$ and all numbers $i\_1,\ldots, i\_t\geq 0$ we have
$$\sum\_{|j\_1|\leq i\_1,\ldots, |j\_t|\leq i\_t}|(A\_1+j\_1)\cap\ldots \cap(A\_t+j\_t)|\leq \sum\_{j\_1|\leq i\_1,\ldots, |j\_t|\leq i\_t}|(I+j\_1)\cap\ldots \cap(I+j\_t)|.$$


Thank you for your attention!



Intersections of translates of finite sets of integers
For the case $t=2$, letting $A\_1:=A$, $A\_2:=-B$, $A\_3:=[-k,k]$, and $A\_4:=[-l,l]$, the sum in the left-hand side counts the number of quadruples $(a\_1,a\_2,a\_3,a\_4)\in A\_1\times A\_2\times A\_3\times A\_4$ with $a\_1+a\_2+a\_3+a\_4=0$.
It is known that this number is maximized, over all quadruples $(A\_1,A\_2,A\_3,A\_4)$ of sets of prescribed odd cardinality, when each set $A\_i$ is a block of consecutive integers centered at $0$. This is a partular case of the rearrangement inequalities due to Gabriel, Hardy, and Littlewood; see [this paper](http://math.haifa.ac.il/seva/Papers/leqfis.dvi) (particularly Theorem 1) for generalizations, references, and the historical background.


This answers your question in the case where $t=2$. For $t>2$, the inequality you are asking about may well be unknown, and yet it should be possible to prove it using the same approach as in the case $t=2$ (basically, by a smart induction, removing simultaneously elements from the sets $A\_i$).


