The answer is $41$. Let $k=n-2$ be the number of triangles, so there are $3k$ internal angles. The total sum of the angles is $180^\circ k$, so $180^\circ k \ge 1^\circ + 2^\circ + \cdots + 3k^\circ\iff 180 \ge \frac{3(k+1)}2\iff k\le 39\iff n\le 41.$ A possible construction is glueing together triangles with internal angles $i^\circ, (i+59)^\circ, (121-2i)^\circ$, $1\le i\le 20$, and $j^\circ, (j+20)^\circ, (160-2j)^\circ$, $21\le j\le 39$. In the first set, $i < i+59 < 121-2i$, in the second set, $j < j+20 < 160-2j$, and for triangles from two different sets, $i < j < j+20 \le 59 < i+59 \le 79 < 82\le 160-2j$, and $121-2i$ and $160-2j$ have different parity.