Romanian Mathematical Olympiad - Final Round

We claim the largest number is $k = 5$.

The difference of two consecutive terms is:
$$\sin(n+i+1)-\sin(n+i) = 2 \sin \frac{1}{2} \cos \frac{2n+2i+1}{2} > 0 \Leftrightarrow \cos \frac{2n+2i+1}{2} > 0. \ (\star)$$

Suppose that for $k = 6$ there exists a positive integer $n$ such that $\sin(n + 1) < \sin(n + 2) < \sin(n + 3) < \sin(n + 4) < \sin(n + 5) < \sin(n + 6)$. Relation $(\star)$ is equivalent to:
$$\cos \left( n + \frac{2i + 1}{2} \right) > 0, \quad i = \overline{1, 5},$$
which implies that:
$$\cos \left(n + \frac{3}{2}\right) + \cos \left(n + \frac{11}{2}\right) = 2 \cos 2 \cdot \cos \left(n + \frac{7}{2}\right) > 0 \Rightarrow \cos 2 > 0,$$
which is a contradiction. Therefore, $k \le 5$.

For $k = 5$ we need to find a positive integer $n$ which, according to $(\star)$, needs to satisfy $\cos(n + \frac{2i+1}{2}) > 0$, $i = \overline{1,4}$. From $2m\pi - \frac{\pi}{2} < n + \frac{3}{2} < n + \frac{9}{2} < 2m\pi + \frac{\pi}{2} \Leftrightarrow (4m-1) \cdot 10\pi < 20n + 30 < 20n + 90 < (4m+1) \cdot 10\pi$, considering $3.14 < \pi < 3.15$, we obtain the following sufficient condition:
$$(4m - 1) \cdot 31.5 < 20n + 30 < 20n + 90 < (4m + 1) \cdot 31.4 \Rightarrow m \le 7.$$
For $m = 7$ we have the following bounds for $n$:
$$\frac{27\pi - 3}{2} < n < \frac{29\pi - 9}{2},$$
which, due to having $3.141 < \pi < 3.142$, implies $n = 41$, and thus proves our claim.