smc

Suppose that such a polygon has $n$ sides. Let the three obtuse angle measures, in degrees, be $o_1$, $o_2$, and $o_3$ and the $(n-3)$ acute angle measures, again in degrees, be $a_1,a_2,a_3,\dots a_{n-3}$. Since $90<o_i<180$ for each $i$, we have $$3(90)=270<o_1+o_2+o_3<3(180)=540,$$ and similarly, since $0<a_i<90$ for each $i$, $$0<a_1+a_2+a_3+\cdots +a_{n-3}<90(n-3)=90n-270.$$ It follows that $$270+0<o_1+o_2+o_3+a_1+a_2+a_3+\cdots +a_{n-3}<540+90n-270,$$ and recalling that the sum of the interior angle measures of an $n$-gon is $180(n-2)=180n-360$, this reduces to $270<180n-360<90n+270$. Hence $$\frac{540}{180}<n<\frac{270+360}{90}⟺3<n<7,$$ so an upper bound is $n\le 6$, and it is easy to check that this bound can be attained by e.g. a convex hexagon with a right angle, $2$ acute angles, and $3$ obtuse angles, as shown below: Accordingly, the maximum possible number of sides of such a polygon is $\boxed{6}$.