Reversed size dependence in ion solubility
This can be explained with some crude thermodynamic models. The lattice enthalpy $\Delta H_\mathrm L$ of an alkali metal halide $\ce{MX}$ is given by

$$\Delta H_\mathrm L = \frac{k_1}{r_\ce{M} + r_\ce{X}}$$

and the hydration enthalpies by

$$\Delta H_\mathrm S(\ce{M}) = -\frac{k_2}{r_\ce{M}}\qquad  \qquad \Delta H_\mathrm S(\ce{X}) = -\frac{k_2}{r_\ce{X}}$$

where $k_1$ and $k_2$ are some positive constants and $r_\ce{M}$ and $r_\ce{X}$ are the ionic radii of $\ce{M+}$ and $\ce{X-}$ respectively.

The solubility is therefore a balance between these two quantities: larger (magnitudes of) solvation enthalpies increase solubility, but larger lattice enthalpies reduce solubility. The solubility $S$ is thus related to the difference of these terms. We ignore the entropic contribution, or to be precise, we assume that $\Delta S$ does not vary too much between different salts $\ce{MX}$.

$$S \sim \frac{k_2}{r_\ce{M}} + \frac{k_2}{r_\ce{X}} - \frac{k_1}{r_\ce{M} + r_\ce{X}}$$

Let us now assume that

$$r_\ce{M} = a+b \qquad \qquad r_\ce{X} = a - b$$

where $a,b$ are more positive constants. Here we have implicitly assumed that $r_\ce{M} \geq r_\ce{X}$, but it works the other way round too, so there is no loss of generality. Thus

$$\begin{align}
S &amp;\sim \frac{k_2}{a+b} + \frac{k_2}{a-b} - \frac{k_1}{2a} \\
&amp;= \frac{2ak_2}{a^2-b^2} - \frac{k_1}{2a}
\end{align}$$

This is clearly a minimum when $b = 0$, i.e. $r_\ce{M} = r_\ce{X}$. So, salts with similar cationic and anionic radii are less soluble, and vice versa.