Find all non-integer solutions of the equation
$$
x+\frac{13}{x}=[x]+\frac{13}{[x]}
$$

where $[x]$ denotes the greatest integer less than or equal to the real number $x$.
(Proposed by the Problem Committee)
The original equation can be transformed into
$$
(x-[x])\left(1-\frac{13}{x[x]}\right)=0 \text {. }
$$

Since $x \notin \mathbf{Z}$, we have $x-[x] \neq 0$. Therefore,
$1-\frac{13}{x[x]}=0 \Rightarrow x[x]=13$.
Let $x=[x]+r(0<r<1)$. Then,
$$
[x]^{2}+r[x]=13 \quad \text{(1)}
$$

When $[x]>0$, we have
$$
[x]^{2}<13<[x]([x]+1) \text {, }
$$

which has no solution.
When $[x]<0$, we have
$$
[x]([x]+1)<13<[x]^{2} \Rightarrow[x]=-4 \text {. }
$$

Substituting into equation (1) gives $x=-\frac{13}{4}$.