Given positive integers $a$, $b$, and $c$ satisfy $a^{2}+b^{2}-c^{2}=2018$.
Find the minimum values of $a+b-c$ and $a+b+c$.
From the problem, we know
$$
a^{2}+b^{2}=c^{2}+2018 \equiv 2 \text { or } 3(\bmod 4)
$$
$\Rightarrow a, b$ are odd, $c$ is even
$$
\begin{array}{l}
\Rightarrow c^{2}+2018=a^{2}+b^{2} \equiv 2(\bmod 8) \\
\Rightarrow c^{2} \equiv 0(\bmod 8) \Rightarrow 4 \mid c .
\end{array}
$$

Notice,
$$
(a+b)^{2}-c^{2}>a^{2}+b^{2}-c^{2}=2018>0 \text {. }
$$

Therefore, $a+b-c>0$.
Since $a+b-c$ is even, thus, $a+b-c \geqslant 2$.
And $1+1009^{2}-1008^{2}=2018$, at this time,
$$
a+b-c=1+1009-1008=2 \text {. }
$$

Thus, the minimum value of $a+b-c$ is 2.
(1) If $c=4$, then
$$
\begin{array}{l}
(a+b)^{2}>a^{2}+b^{2}=c^{2}+2018>45^{2} \\
\Rightarrow a+b \geqslant 46 ;
\end{array}
$$

If $a+b=46$, then
$$
a^{2}+b^{2}=a^{2}+(46-a)^{2}=2034
$$
$\Rightarrow a^{2}-46 a+41=0$, this equation has no integer solutions;
If $a+b=48$, then
$$
\begin{array}{l}
a^{2}+b^{2}=a^{2}+(48-a)^{2}=2034 \\
\Rightarrow a^{2}-48 a+135=0 \Rightarrow a=3 \text { or } 45 \\
\Rightarrow a+b+c=52 ;
\end{array}
$$

If $a+b \geqslant 50$, then $a+b+c \geqslant 54$.
(2) If $c \geqslant 8$, similarly, $a+b \geqslant 46$, hence
$$
a+b+c \geqslant 46+8=54 \text {. }
$$

In summary, the minimum value of $a+b-c$ is $2$, and the minimum value of $a+b+c$ is 52.