It is easy to show that equality holds when $m=\frac{1}{\sin(n+1)}.$ For example, working from the last term to the first, we get $\frac{1}{\sin(n+1)}\cdot\frac{\sin n}{\sin^2 1}-\frac{1}{\sin n\cdot\sin(n+1)}=\frac{1}{\sin(n+1)}\cdot\frac{(\sin n-\sin 1)(\sin n+\sin 1)}{\sin n\cdot\sin^2 1}=\frac{\sin(n-1)}{\sin n\cdot \sin^2 1},$ and so on. Therefore, if $m=1,$ we get $\sin(n+1)=1,$ and thus $n=89,449,\cdots ,10529.$ If $m=2,$ we get $\sin(n+1)=\frac{1}{2},$ and thus either $n=29, 389, \cdots, 10469,$ or $n=149, 509, \cdots 10589.$ It is easy to show that the three arithmetic sequences above do not have a common term. As such, each sequence having 30 terms, we have a total of 90 acceptable values of $n.$

N.T.TUAN wrote: Let $m$ be equal to $1$ or $2$ and $n<10799$ be a positive integer. Determine all such $n$ for which $\sum_{k=1}^{n}\frac{1}{\sin{k}\sin{(k+1)}}=m\frac{\sin{n}}{\sin^{2}{1}}$. $\sum_{k=1}^{n}\frac{1}{\sin{k}\sin{(k+1)}}$ $=\frac{1}{\sin 1}\sum_{k=1}^{n}\frac{\sin 1}{\sin{k}\sin{(k+1)}}$ $=\frac{1}{\sin 1}\sum_{k=1}^{n}\frac{\sin (k+1-k)}{\sin{k}\sin{(k+1)}}$ $=\frac{1}{\sin 1}\sum_{k=1}^{n}\frac{\sin (k+1)\cos k-\cos(k+1)\sin k}{\sin{k}\sin{(k+1)}}$ $=\frac{1}{\sin 1}\sum_{k=1}^{n}\left(\cot k-\cot (k+1)\right)$ $=\frac{\cot 1-\cot(n+1)}{\sin 1}$ $=\frac{\sin n}{\sin^21\sin(n+1)}$ $m=\frac{1}{\sin(n+1)}$ $\sin(n+1)=1$ or $\sin(n+1)=\frac{1}{2}$ Now,the values of n can be easily computed.

The truth is, one would be hard-pressed to guess what $m$ is, and the statement of the problem makes it seem much harder than it is. In reality, I would guess that most people will try the first couple cases of $n=1$ and $n=2,$ realize what $m$ should be, and then try to prove it in general. Once you get there, the simplest and most direct way (and nothing against pankajsinha's approach, which is very nice) is to use induction. That's what I did but I tried to keep the typing to a minimum so I included only the last step of the induction process.

Solving further $n+1=\frac{(4k+1)\pi}{2}$ or $n+1=p\pi+(-1)^p\frac{\pi}{6}$ $k$ and $p$ being integers. In both the cases,$n$ is an irrational number and not a positive integer as claimed in the question and $n$ being a positive integer is also some kind of requirement since the question is in terms of $\sum_{k=1}^n$. So answer should be that for no value of $n$ the statement $ \sum_{k=1}^{n}\frac{1}{\sin{k}\sin{(k+1)}}=m\frac{\sin{n}}{\sin^{2}{1}} $ is possible.

Except that these numbers are in degrees, not radians... At least that's how I understood the intent...

If it's in degrees then the sum is only defined for $0<n<179$.