Edit: here is a new version, with lighter computations and, hopefully, no error. My thanks to Did for pointing out some problems in my former proof. Of course, my answer is now essentially the same as Did...
Let $(X,Y)$ be a couple of random variables uniformly distributed in $[0,1]^2$. Let $\delta \in [0,1)$. The event $\{|X-Y|>\delta\}$ has probability:
$\int_{[0,1]^2} 1_{|x-y|> \delta} \ dx \ dy = 2 \int_{[0,1]} \int_{[0,1]} 1_{x > \delta + y} dx \ dy = (1-\delta)^2.$
Let $(\tilde{X}, \tilde{Y})$ be the random variables $(X,Y)$ conditioned by the event $\{|X-Y|>\delta\}$. It has the following density with respect to the Lebesgue measure on $[0,1]^2$:
$\frac{1_{x > y + \delta} + 1_{x < y - \delta}}{(1-\delta)^2}.$
Hence:
$\mathbb{E} ((\tilde{X}-\tilde{Y})(1-2\tilde{X})) = \frac{1}{(1-\delta)^2} \int_{[0,1]^2} (x-y)(1-2x) (1_{x > y + \delta} + 1_{x < y - \delta}) \ dx \ dy$
$\cdots = \frac{1}{(1-\delta)^2} \int_\delta^1 \int_0^{x-\delta} (x-y)(1-2x) \ dy \ dx + \frac{1}{(1-\delta)^2} \int_0^{1-\delta} \int_{x+\delta}^1 (x-y)(1-2x) \ dy \ dx $
The change of variables $(u,v) = (1-x,1-y)$ shows that these two integrals are the same, so we only need to compute one of them. Thanks to Wolfram,
$\int_\delta^1 \int_0^{x-\delta} (x-y)(1-2x) \ dy \ dx = -\frac{1}{2} \int_\delta^1 (2x-1) (x^2-\delta^2) \ dx = -\frac{(1-\delta)^2 (1+2\delta+3\delta^2)}{12}.$
Thus:
$\mathbb{E} ((\tilde{X}-\tilde{Y})(1-2\tilde{X})) = -\frac{1+2\delta+3\delta^2}{6}.$
This formula gives the good limits when $\delta$ goes to $0$ or $1$. It is also always negative (even for $\delta = 0$). By the law of large numbers, you should expect the sum to be negative for large $n$ almost surely.