Answer: 197

Small Hint:

The centers sit at height $\rho$ above $BC,$ spaced $2\rho$ apart, and the end centers lie at distances $\rho\cot\frac{B}{2}$ and $\rho\cot\frac{C}{2}$ from $B$ and $C$

Big Hint:

Both chains measure the same $BC,$ which determines $\cot\frac{B}{2} + \cot\frac{C}{2};$ the incircle satisfies $BC = r\left(\cot\frac{B}{2} + \cot\frac{C}{2}\right)$

Solution:

For a chain of $n$ circles of radius $\rho$ tangent to $\overline{BC},$ the centers lie at height $\rho$ with consecutive centers $2\rho$ apart. The first circle is tangent to $\overline{AB}$ and $\overline{BC},$ so its center lies on the bisector from $B,$ at horizontal distance $\rho\cot\frac{B}{2}$ from $B;$ similarly the last center is $\rho\cot\frac{C}{2}$ from $C.$ Hence with $k = \cot\frac{B}{2} + \cot\frac{C}{2},$ $$BC = \rho k + 2\rho(n - 1).$$

The two chains give $34k + 34 \cdot 14 = BC$ $= k + 2 \cdot 2023,$ so $33k = 3570$ and $k = \frac{1190}{11},$ whence $BC = k + 4046 = \frac{45696}{11}.$

The incircle is a chain of one circle of radius $r:$ $BC = rk.$ Therefore $$r = \frac{BC}{k} = \frac{45696}{1190} = \frac{192}{5},$$ and $m + n = 192 + 5 = 197.$