Solution:
By rationalizing the denominator, $\frac{a \sqrt{2013}+b}{b \sqrt{2013}+c}=\frac{2013 a b-b c+\sqrt{2013}\left(b^{2}-a c\right)}{2013 b^{2}-c^{2}}$. Since this is rational, then $b^{2}-a c=0$. Consequently,
$$\begin{aligned} a^{2}+b^{2}+c^{2} & =a^{2}+a c+c^{2}=(a+c)^{2}-a c=(a+c)^{2}-b^{2} \\ & =(a-b+c)(a+b+c) \end{aligned}$$
and
$$\begin{aligned} a^{3}-2 b^{3}+c^{3} & =a^{3}+b^{3}+c^{3}-3 b^{3}=a^{3}+b^{3}+c^{3}-3 a b c \\ & =(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \end{aligned}$$
Therefore,
$$\frac{a^{2}+b^{2}+c^{2}}{a+b+c}=a-b+c \quad \text{ and } \quad \frac{a^{3}-2 b^{3}+c^{3}}{a+b+c}=a^{2}+b^{2}+c^{2}-a b-b c-c a$$
are integers.