How many sides can a convex polygon have if the sum of its obtuse angles is $3000^{\circ}$?
Note that a convex polygon cannot have more than three non-obtuse angles (unless it is a rectangle). Indeed, if there are more than three such angles, then the external angles adjacent to them are not acute, which contradicts the fact that the sum of the external angles of a convex polygon (taken one at each vertex) is $360^{\circ}$.

Thus, the sum of the angles of the given polygon is $3000^{\circ}+S^{\circ}$, where $0 \leq S \leq 270$. Let it have $n$ sides, then $180(n-2)=3000+S$, that is,

$n=2+(3000+S): 180 \Leftrightarrow n=18+(120+S): 180$. Considering that $n$ is a natural number, we get: $S=60 ; n=$ 19 or $S=240, n=20$.

A 19-sided polygon, where one acute angle is $60^{\circ}$ and 18 obtuse angles with a given sum, as well as a 20-sided polygon with three acute angles, the sum of which is $240^{\circ}$, and 17 obtuse angles with a given sum, is not difficult to "construct". For example, in the first case, we sequentially lay out 19 chords on a circle, corresponding to a central angle of $13^{1 / 3}{ }^{\circ}$. Extending the ends of the extreme chords until they intersect, we obtain a 19-sided polygon, where 18 angles are equal to $166^{2} / 3^{\circ}$.

## Answer

19 or 20.