Does anybody happen to know the solution to this problem? In particular it doesn't appear to be easy for even n=4.

The failure of SSA similarity is the key here. A sketch: Suppose we have an obtuse triangle $ABC$ with rational sides, obtuse angle at $C$, and shortest side $BC$. If we replace $C$ with its reflection over the altitude from $A$, the new triangle has the same altitude and rational sides. You can determine how many times this construction can be iterated by looking at angles.

MellowMelon's post outlines the idea that solves the problem, but here's some more detail: First replace the word "integer" in the problem with the word "rational" (we can do this by scaling each triangle). Lemma 1 (MellowMelon): Suppose we have an obtuse triangle $ABC$ with rational sides, obtuse angle at $C$, and shortest side $BC$. If we replace $C$ with its reflection over the altitude from $A$, the new triangle has the same altitude and rational sides. Proof: Let $ c = AB, b = CA, a = BC, $ and let $ d $ be the length of the altitude from $ A. $ Also let $ Z $ be the foot of the $ A $-altitude on $ BC $ and let $ x = CZ. $ It is clear that $ d^2 + (a + x)^2 = c^2 $ and $ d^2 + x^2 = b^2 $ so by subtracting these equations we find that $ x = \frac{c^2 - a^2 - b^2}{2a} \in \mathbb{Q} $ as desired. Now we return to the problem. Suppose we have an obtuse triangle with smallest angle $ \alpha $ and second smallest angle $ \beta. $ Clearly the obtuse angle has measure $ 180 - \alpha - \beta. $ Upon performing the operation described in Lemma 1 we find that the new triangle has an angle with measure $ 180 - \alpha - 2\beta > 180 - 2(\alpha + \beta). $ We can perform this operation so long as the triangle we have is obtuse (and it is clear that if after the operation we obtain a new, obtuse triangle, the "new" side length is the longest side of the triangle. Therefore by forcing $ \alpha + \beta $ to be sufficiently small (which is equivalent to the original obtuse angle being sufficiently large), we can perform arbitrarily many operations. So to solve the problem, it suffices to find a triangle with rational side lengths and rational area whose largest angle is arbitrarily close to $ 180 $ degrees. But this isn't hard - consider the triangle with side lengths $ 8(m + 1)^2 $ and $ ((2m + 1)^2 + 1)(2m + 3) $ and $ ((2m + 3)^2 + 1)(2m + 1). $ The altitude to the shortest side's length can be computed to be $ 2(2m + 1)(2m + 3). $ It is clear that as we let $ m $ be a sufficiently large rational number, our triangle becomes "arbitrarily obtuse" so we are done.

Wolstenholme wrote: MellowMelon's post outlines the idea that solves the problem, but here's some more detail: First replace the word "integer" in the problem with the word "rational" (we can do this by scaling each triangle). Lemma 1 (MellowMelon): Suppose we have an obtuse triangle $ABC$ with rational sides, obtuse angle at $C$, and shortest side $BC$. If we replace $C$ with its reflection over the altitude from $A$, the new triangle has the same altitude and rational sides. Proof: Let $ c = AB, b = CA, a = BC, $ and let $ d $ be the length of the altitude from $ A. $ Also let $ Z $ be the foot of the $ A $-altitude on $ BC $ and let $ x = CZ. $ It is clear that $ d^2 + (a + x)^2 = c^2 $ and $ d^2 + x^2 = b^2 $ so by subtracting these equations we find that $ x = \frac{c^2 - a^2 - b^2}{2a} \in \mathbb{Q} $ as desired. Now we return to the problem. Suppose we have an obtuse triangle with smallest angle $ \alpha $ and second smallest angle $ \beta. $ Clearly the obtuse angle has measure $ 180 - \alpha - \beta. $ Upon performing the operation described in Lemma 1 we find that the new triangle has an angle with measure $ 180 - \alpha - 2\beta > 180 - 2(\alpha + \beta). $ We can perform this operation so long as the triangle we have is obtuse (and it is clear that if after the operation we obtain a new, obtuse triangle, the "new" side length is the longest side of the triangle. Therefore by forcing $ \alpha + \beta $ to be sufficiently small (which is equivalent to the original obtuse angle being sufficiently large), we can perform arbitrarily many operations. So to solve the problem, it suffices to find a triangle with rational side lengths and rational area whose largest angle is arbitrarily close to $ 180 $ degrees. But this isn't hard - consider the triangle with side lengths $ 8(m + 1)^2 $ and $ ((2m + 1)^2 + 1)(2m + 3) $ and $ ((2m + 3)^2 + 1)(2m + 1). $ The altitude to the shortest side's length can be computed to be $ 2(2m + 1)(2m + 3). $ It is clear that as we let $ m $ be a sufficiently large rational number, our triangle becomes "arbitrarily obtuse" so we are done. Could you tell me the way you choose three sides of the triangle ( $ 8(m + 1)^2 $ and $ ((2m + 1)^2 + 1)(2m + 3) $ and $ ((2m + 3)^2 + 1)(2m + 1) $), please ?????

I think a reasonable way to find out the construction is to make the length of three side satisfying $x+y-z=2$. To make the area is an integer, we may use Heron formula, which implies $(x-1)(y-1)(x+y-1)$ is square. An intuitive thought is to make $x-1=s^2, y-1=t^2$. Hence $s^2+t^2+1$ is a square, and we can make $s=2n^2, t=2n$