The required minimum value is 2019.
$$
\text { Let } x^{2}+y^{2}-2017 x y=k \text {, }
$$

where $k$ is not a perfect square.
First, prove that $: \geqslant 2019$.
Let $\left(x_{0}, y_{0}\right)$ be the positive integer solution of equation (1) that minimizes $x_{0}+y_{0}$. Without loss of generality, assume $x_{0} \geqslant y_{0}$.
By the condition, $x_{0}$ is a root of the quadratic equation
$$
x^{2}-2017 y_{0} x+y_{0}^{2}-k=0
$$

Let the other root be $x_{1}$. Then
$$
\left\{\begin{array}{l}
x_{0}+x_{1}=2017 y_{0}, \\
x_{0} x_{1}=y_{0}^{2}-k .
\end{array}\right.
$$

From the first equation, we know $x_{1} \in \mathbf{Z}$.
Since $k$ is not a perfect square, then $x_{1} \neq 0$.
If $x_{1}>0$, then
$$
\begin{array}{l}
x_{1}=\frac{y_{0}^{2}-k}{x_{0}}<\frac{y_{0}^{2}}{x_{0}} \leqslant \frac{x_{0}^{2}}{x_{0}}=x_{0} \\
\Rightarrow x_{1}+y_{0}<x_{0}+y_{0},
\end{array}
$$

and $\left(x_{1}, y_{0}\right)$ is also a positive integer solution of equation (1), which contradicts the minimality of $x_{0}+y_{0}$.
Thus, it must be that $x_{1}<0$.
From the system of equations (2), we have
$$
\begin{array}{l}
x_{0} \geqslant 2017 y_{0}+1, y_{0}^{2}-k \leqslant-1 . \\
\text { Hence } k=x_{0}\left(x_{0}-2017 y_{0}\right)+y_{0}^{2} \\
\geqslant\left(2017 y_{0}+1\right) \cdot 1+y_{0}^{2} \geqslant 2019 . \\
\text { Let } x^{2}+y^{2}-2017 x y=2019, \text { then } \\
(x+y)^{2}=2019(x y+1) .
\end{array}
$$

Take $x y+1=2019, x+y=2019$.
Solving, we get $(x, y)=(2018,1)$ or $(1,2018)$.
Therefore, the required minimum value is 2019.
(Fang Tinggang, Chengdu No.7 High School, Sichuan Province, 610041)
Let $x, y$ be positive integers,
$$
x^{2}+y^{2}-2017 x y>0
$$

and not a perfect square. Find the minimum value of $x^{2}+y^{2}-2017 x y$.