Solution:

Abbreviate integer-ratio right triangle by IRRT.

The square $(n=4)$ has such a decomposition. For example, a $12 \times 12$ square can be cut into twenty-four 3-4-5 triangles as shown below:

Now we show that $n=4$ is the only solution. The proof is by contradiction. Suppose an $n$-gon with $n \neq 4$ (and $n \geq 3$) has a dissection into IRRTs. Choose an arbitrary vertex $P$ of the $n$-gon. One or more IRRTs meet at $P$; let their internal angles at $P$ be $\theta_{1}, \theta_{2}, \ldots, \theta_{k}$, where we have
$$\theta_{1}+\theta_{2}+\cdots+\theta_{k}=\frac{n-2}{n} \cdot \pi .$$
Observe that $\sin \theta_{i}$ and $\cos \theta_{i}$ are rational for each $i=1, \ldots, k$. By applying angle-sum identities, we may infer that the sine and cosine of $\frac{n-2}{n} \cdot \pi$, and thus of $2 \pi / n$, are rational as well. Applying angle-sum identities again, it follows that for any divisor $m$ of $n$, the sine and cosine of $2 \pi / m$ are also rational. Since $n \neq 4$ and $n \geq 3, n$ has a divisor which is either 8 or an odd prime. So, we can restrict our attention to these two cases.

For $n=8$, we have $\sin (2 \pi / 8)=\sqrt{2} / 2$. This is irrational, so we have our contradiction.

Now we consider the case $n=p$ where $p$ is an odd prime. Let $t=\tan (2 \pi / p)$, which must be rational. By the multiple-angle identity for tangents, $t$ satisfies
$$p t-\left(\begin{array}{l} p \\ 3 \end{array}\right) t^{3}+\left(\begin{array}{c} p \\ 5 \end{array}\right) t^{5}-\cdots \pm t^{p}=0$$
By the Rational Root Theorem, the only possible rational values of $t$ are $t= \pm 1$ and $t= \pm p$. But $t \neq \pm 1$, because $\frac{2 \pi}{p} \neq \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$. If $t= \pm p$, then the left-hand side of (2) is congruent to $p^{2}\left(\bmod p^{3}\right)$ and thus cannot equal 0. Again, we have a contradiction.

It follows that no $n$-gon with $n \neq 4$ has a dissection into IRRTs.