jmc

Let the two different sides of the rectangle be $a$ and $b$. Since the perimeter is 26 units, we have the equation $2a+2b=26\Rightarrow a+b=13$. We can rearrange this equation to get $a=13-b$. We want to minimize the value of $\sqrt{a^2+b^2}$. Substituting in the last equation, we have $\sqrt{(13-b{)}^2+b^2}=\sqrt{169-26b+2b^2}$. This value is minimized when the quadratic $169-26b+2b^2$ is minimized, which occurs when $b$ (which must be an integer) is as close to $-\frac{-26}{2\cdot 2}=6.5$ as possible. So let $b=7$ and $a=13-7=6$. So, the shortest possible diagonal is $\sqrt{7^2+6^2}=\boxed{\sqrt{85}}$.