Write $\sin 1^\circ = \sin\big((k+1)^\circ - k^\circ\big)$ to show $\frac{1}{\sin k^\circ \sin(k+1)^\circ} = \frac{\cot k^\circ - \cot(k+1)^\circ}{\sin 1^\circ}$

Use $\cot(180^\circ - x) = -\cot x:$ the cotangents at supplementary arguments cancel in pairs, leaving only the $45^\circ$ term

Since $\sin 1^\circ = \sin\big((k+1)^\circ - k^\circ\big)$ $= \sin(k+1)^\circ \cos k^\circ$ $- \cos(k+1)^\circ \sin k^\circ,$ dividing by $\sin k^\circ \sin(k+1)^\circ$ gives $$\begin{aligned} &\small \frac{1}{\sin k^\circ \sin(k+1)^\circ} \\ &\scriptsize = \frac{\cot k^\circ - \cot(k+1)^\circ}{\sin 1^\circ}. \end{aligned}$$ So the sum times $\sin 1^\circ$ equals $\cot 45^\circ - \cot 46^\circ + \cot 47^\circ$ $- \cot 48^\circ$ $+ \cdots + \cot 133^\circ - \cot 134^\circ,$ with $+$ signs on odd arguments and $-$ signs on even arguments.

Because $\cot(180^\circ - x) = -\cot x$ and supplementary arguments here have the same parity, the terms cancel in supplementary pairs: $+\cot 133^\circ$ cancels $+\cot 47^\circ,$ $-\cot 134^\circ$ cancels $-\cot 46^\circ,$ and so on for every pair of arguments summing to $180^\circ.$ The only survivors are $\cot 45^\circ = 1$ (its partner $135^\circ$ is out of range) and $-\cot 90^\circ = 0.$

Hence the sum equals $\frac{\cot 45^\circ}{\sin 1^\circ} = \frac{1}{\sin 1^\circ},$ so the least such $n$ is $1.$