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The answer is $n=3,4,5,7$. We first introduce a variant of the $k$th Chebyshev polynomials in the following lemma (which is standard, and easily shown by induction).

Lemma For each $k\ge 0$ there exists $P_k(X)\in \mathbb{Z}[X]$, monic for $k\ge 1$ and with degree $k$, such that $$P_k(X+X^{-1})\equiv X^k+X^{-k}.$$ The first few are $P_0(X)\equiv 2$, $P_1(X)\equiv X$, $P_2(X)\equiv X^2-2$, $P_3(X)\equiv X^3-3X$.

Suppose the angles of the triangle are $\alpha <\beta <\gamma$, so the law of cosines implies that $$2\cos \alpha =\frac{n+4}{n+1}\text{and}2\cos \gamma =\frac{n-4}{n-1}.$$

Claim — The triangle is quirky iff there exists $r,s\in {\mathbb{Z}}_{\ge 0}$ not both zero such that $$\cos (r\alpha )=\pm \cos (s\gamma )\text{or equivalently}{P}_r\left(\frac{n+4}{n+1}\right)=\pm P_s\left(\frac{n-4}{n-1}\right).$$ Proof. If there are integers $x,y,z$ for which $x\alpha +y\beta +z\gamma =0$, then we have that $(x-y)\alpha =(y-z)\gamma -\pi y$, whence it follows that we may take $r=|x-y|$ and $s=|y-z|$ (noting $r=s=0$ implies the absurd $x=y=z$). Conversely, given such $r$ and $s$ with $\cos (r\alpha )=\pm \cos (s\gamma )$, then it follows that $r\alpha \pm s\gamma =k\pi =k(\alpha +\beta +\gamma )$ for some $k$, so the triangle is quirky. $\square$

If $r=0$, then by rational root theorem on $P_s(X)\pm 2$ it follows $\frac{n-4}{n-1}$ must be an integer which occurs only when $n=4$ (recall $n\ge 3$). Similarly we may discard the case $s=0$.

Thus in what follows assume $n\ne 4$ and $r,s>0$. Then, from the fact that $P_r$ and $P_s$ are nonconstant monic polynomials, we find

Corollary If $n\ne 4$ works, then when $\frac{n+4}{n+1}$ and $\frac{n-4}{n-1}$ are written as fractions in lowest terms, the denominators have the same set of prime factors.

But $\gcd (n+1,n-1)$ divides 2, and $\gcd (n+4,n+1)$, $\gcd (n-4,n-1)$ divide 3. So we only have three possibilities:

$n+1=2^u$ and $n-1=2^v$ for some $u,v\ge 0$. This is only possible if $n=3$. Here $2\cos \alpha =\frac{7}{4}$ and $2\cos \gamma =-\frac{1}{2}$, and indeed $P_2(-1/2)=-7/4$.

$n+1=3\cdot 2^u$ and $n-1=2^v$ for some $u,v\ge 0$, which implies $n=5$. Here $2\cos \alpha =\frac{3}{2}$ and $2\cos \gamma =\frac{1}{4}$, and indeed $P_2(3/2)=1/4$.

* $n+1=2^u$ and $n-1=3\cdot 2^v$ for some $u,v\ge 0$, which implies $n=7$. Here $2\cos \alpha =\frac{11}{8}$ and $2\cos \gamma =\frac{1}{2}$, and indeed $P_3(1/2)=-11/8$.

Finally, $n=4$ works because the triangle is right, completing the solution.