For $p\in[0,1]$, we write $\mathrm{Ber}(p)$
to denote the Bernoulli measure on $\{0,1\}$;
that is, $\mathrm{Ber}(p)(0)=1-p$,
$\mathrm{Ber}(p)(1)=p$.
For $n\in\mathbb{N}$ and $p=(p_1,\ldots,p_n)\in[0,1]^n$, 
$\mathrm{Ber}(p)$ denotes
the product of $n$ Bernoulli distributions with parameters $p_i$:
$$
\mathrm{Ber}(p) =
\mathrm{Ber}(p_1) \otimes \mathrm{Ber}(p_2)\otimes \ldots \otimes \mathrm{Ber}(p_n).
$$
Thus, $\mathrm{Ber}(p)$ is a probability measure on 
$\{0,1\}^n$,
with
$$
\mathrm{Ber}(p)(x) 
= \prod_{i=1}^n p_i^{x_i}(1-p_i)^{1-x_i}
,
\qquad
x\in\{0,1\}^n,
$$

Consider the following (non-convex*) optimization problem:
Minimize 
$$
||\mathrm{Ber}(p)
-
\mathrm{Ber}(q)||_1
$$
over all $p,q\in[0,1]^n$
under the constraints $p_i-q_i=\varepsilon_i\ge0$
and $\sum_{i=1}^n \varepsilon_i=\ell$.

I am only interested in the case of even $n$.
I thought I could prove, but now only conjecture
(with *very* compelling numerical evidence)
that the unique minimum
occurs at $p_i=1/2+\ell/(2n)$
and $q_i=1/2-\ell/(2n)$.

How does one prove this?

Warning: the above is only optimal for even $n$, not odd.


*I had previously wrongly claimed that the problem was convex.