It is possible for all integer $N \ge 3$
It is obvious that a triangle can't be partitioned in $1$ quadrilateral.
Also if we divide a triangle into $2$ quadrilaterals, one is convex but other one is not.
For $N=3$, take a equilateral triangle. And divide the triangle into $3$ congruent quadrilaterals. Like the figure. $O$ is the circumcenter. We draw $\angle OFB$ such that it is equal to $60^o$. Also same thing for $ \angle ODC $ and $ \angle OEA$.
It is easy to prove that the quadrilaterals are congruent.
Now we can make a quadrilateral $BCYX$ such that it is similar to $EOFB$. And $\angle ABX = \angle ACY = 60^o+120^o=180^o$. So that means $\triangle AXY$ is a new triangle with $5$ similar quadrilaterals. We can make this process over and over. So it is possible for all $N \ge 3$
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