Let $\tau>0$ and $\mathcal{M}$ be the set of absolutely continuous cumulative distribution functions. I am interested in solving:
\begin{align} \underset{F \in \mathcal{M}}{\text{arg min}} \int_{\mathbb{R}}F(x+\tau)F'(x)dx \end{align} The problem might require some moment constraints on the distribution in $\mathcal{M}$. Indeed for $F \in \mathcal{M}$, since $\tau>0$ we have: \begin{align} \int_{\mathbb{R}}F(x+\tau)F'(x)dx > \int_{\mathbb{R}}F(x)F'(x)dx = \frac{1}{2}. \end{align} Moreover letting $F$ be the c.d.f. of $\mathcal{N}(0,\sigma^2)$ and $\Phi$, $\phi$ be the c.d.f. and p.d.f of $\mathcal{N}(0,1)$ respectively we have: \begin{align} \int_{\mathbb{R}}F(x+\tau)F'(x)dx &= \int_{\mathbb{R}}\Phi\left(\frac{x+\tau}{\sigma}\right)\frac{1}{\sigma}\phi\left(\frac{x}{\sigma}\right)dx \\ &= \int_{\mathbb{R}}\Phi\left(u+\frac{\tau}{\sigma}\right)\phi(u)du \end{align} and by the dominated convergence theorem we conclude: \begin{align} \int_{\mathbb{R}}F(x+\tau)F'(x)dx \rightarrow \frac{1}{2} \quad \text{as $\sigma \rightarrow \infty$.} \end{align} Hence the minimum is not attained. Assuming for instance that $\mathcal{M}$ contains only c.d.f. with mean 0 and variance 1, is there a solution to the problem? For small $\tau$, it is easily shown that:
\begin{align} \int_{\mathbb{R}}F(x+\tau)F'(x)dx \underset{\tau \rightarrow 0}{\sim} \frac{1}{2} + \tau\int_{\mathbb{R}}F'(x)^2dx \end{align} hence the temptation to select the distribution with minimum $L_2$ norm. But $\tau$ may be arbitrarily large...
What about if $\mathcal{M}$ was restrained to distributions supported over $[a,b]$, $-\infty<a<b<\infty$ ?
Suppose that $F(x)=P(U\le x)$ where $U$ is a random variable with the uniform distribution over $[-a,a]$. Then, $F(x+\tau) = 1$ if $x\ge a-\tau$ and $F(x+\tau) = \frac{x+\tau-a}{2a}$ if $-a-\tau < x < a-\tau$, and \begin{align} \int_{\mathbb{R}}F(x+\tau)F'(x)dx % & = \frac{1}{2a}\int_{-a}^{a}F(x+\tau) dx \\ % & = \frac{1}{2a} \left( \int_{-a}^{a-\tau}\frac{x+\tau-a}{2a} dx + \int_{a-\tau}^{a} dx\right) \\ % & = \frac{1}{2a} \left( \int_{-a}^{a-\tau}\frac{x}{2a} dx + \frac{\tau-a}{2a}(2a-\tau) + \tau \right) \\ % & = \frac{1}{2a} \left( \frac{\tau^2 - 2a\tau}{4a} + \frac{\tau-a}{2a}(2a-\tau) + \tau \right) \\ % & = \frac{1}{2a} \left( \frac{-\tau^2 }{4a} + 2\tau-a \right) . \end{align} Now, \begin{equation} \frac{1}{2a} \left( \frac{-\tau^2 }{4a} + 2\tau-a \right) = \frac{1}{2} \end{equation} if and only if \begin{equation} % 8a^2- 4\tau a - \tau^2 = 0 % a^2 - a\tau + \frac{\tau^2}{8} = 0 % \end{equation} Taking \begin{equation} a= \tau \frac{ 1 \pm \frac{\sqrt{2}}{2} } {2} \end{equation} you obtain two $F$s that match your lower bound. Does this make sense?