$4 \cdot \frac{41}{37}$. For (1), by the theorem of the ratio of sums and differences, we get
$$\begin{aligned} \frac{m+a n}{b m+c n} & =\frac{(x+9 y)+a(9 x+7 y)}{b(x+9 y)+c(9 x+7 y)} \\ & =\frac{(1+9 a) x+(9+7 a) y}{(b+9 c) x+(9 b+7 c) y} . \end{aligned}$$

From (2), we have $\frac{x+9 y}{9 x+8 y}=\frac{(1+9 a) x+(9+7 a) y}{(b+9 c) x+(9 b+7 c) y}$.
The above equation holds for all real numbers $x, y, m, n$ if and only if there exists a non-zero constant $k$ such that
$$\left\{\begin{array}{l} 1+9 a=k, \\ 9+7 a=9 k, \\ b+9 c=9 k, \\ 9 b+7 c=8 k . \end{array}\right.$$

From the first two equations, we get $a=0, k=1$; thus, from the last two equations, we solve for $b$
$$\begin{array}{l} =\frac{9}{74}, c=\frac{73}{74} . \\ \quad \therefore a+b+c=\frac{82}{74}=\frac{41}{37} . \end{array}$$