Let $x=a+k$, where $a \in \mathbb{Z}$ and $0<k<1$. Here, we have, \[ a+k+\frac{13}{a+k} = a+\frac{13}{a} \] It is equivalent to $k=\frac{13}{a} - a$. Since $0<k<1$, thus $a^2<13$ and $a^2+a-13>0$. Since $a \in \mathbb{Z}$, hence $a^2 \leq 9$ and $a \leq 3$. But, then $-1 \geq a^2+a-13>0$. Thus, there is no solution to this equation.

And then, there's also such thing as negative integers. If $k={13\over a}-a$ and $a<0$, then $k>0\implies a^2>13\implies a\leqslant -4$ and $k<1\implies a^2+a-13<0\implies -4\leqslant a\leqslant 3$. Hence the only possible value for $a$ is $a=-4$, yielding $x=a+k={13\over a}=\boxed{-{13\over 4}}$