Just use right-angled triangles with side lengths 1, 2, and root(3). Show that if it's possible for n, it's possible for n+2 (just add 2 triangles which form a rectangle in the middle). Then we are only left to show n=6 and n=7 are possible. Easy problem.

I wasn't able to follow al.M.V.'s sketch. I'm sorry if my solution is very similar to his one. For $ n=6$ take a regular hexagon divided into $ 6$ equilateral triangles. Check the image. The first hexagon is made up of $ 4k$ congruent right triangles. The second one is made up of $ 4k+2$ congruent right triangles, where $ k\geq2$ is equal to $ 1+$length of horizontal side of the hexagon. For odd $ n$ we use right rectangles with sides $ 1,3$ and $ \sqrt{10}$ respectively, arranged like in the third picture. If $ k$ is the length of the right vertical side, then we have a hexagon made up of $ 4+2k+1=2k+5\geq7$ congruent triangles. Of course, it is obvious that each small rectangle consists of 2 right triangles. Invalid image file Just in case the above image doesn't show up, the attached image is also available:

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al.M.V. wrote: Just use right-angled triangles with side lengths 1, 2, and root(3). Show that if it's possible for n, it's possible for n+2 (just add 2 triangles which form a rectangle in the middle). Then we are only left to show n=6 and n=7 are possible. Easy problem. I think your solution can't work when n is odd.Can you explain it?

Look at the third diagram. It is possible to write triangles with such lengths.