Assume the sidelengths are $l_1,...,l_n$ in order, so that $\{l_1,...,l_n\}=\{1,...,n\}$. Here is the main claim: If (and only if) we can find $x_1,...,x_n>0$ such that $l_i=x_i+x_{i+1}$ (indices $\pmod n$) for all $i$, then there exist a circumscribable polygon with sidelengths $l_1,...,l_n$. Consider the function $f(r)=2\sum_{i=1}^n \arctan(\frac{x_i}{r})$. Since $\lim_{r\rightarrow 0^+}f(r)=n\pi$ and $\lim_{r\rightarrow +\infty}f(r)=0$, it follows by continuity that there is a value such that $f(r_0)=2\pi$. This means that there exists a circumscribed polygon of inradius $r_0$, with the tangency segments being the $x_i$s, which closes itself. If $n=4k+2$, then the condition implies $l_1-l_2+l_3-\cdots+l_{n-1}-l_n=0$. However, taking $\pmod 2$, we have $0=l_1-\cdots -l_n\equiv l_1+\cdots +l_n=1+\cdots +n=\frac{n(n+1)}{2}=\frac{n}{2}(n+1)\equiv 1\pmod{2}$ which is a contradiction. If $n$ is odd, pick $(l_1,l_2,...,l_n)=(1,...,n)$, and pick any $x_1\in (0,1)$. Then by induction $x_k=(k-1)-(k-2)+(k-3)-\cdots +(-1)^k1+(-1)^{k+1}x_1$. If $k$ is odd, $x_k\geq \frac{k-1}{2}+x_1>0$, and if $k$ is even $x_k\geq \frac{k-2}{2}+1-x_1>0+0=0$. If $n=4t$, pick $(l_1,...,l_n)=(1,2,4,3,5,6,8,7,...,4i+1,4i+2,4i+4,4i+3,...,4k-3,4k-2,4k,4k-1)$. Once again pick $x_1\in (0,1)$. By induction $x_{4i+1}=(4i-1)-4i+(4i-2)-(4i-3)+x_{4i-3}=x_{4i-3}=x_{4i-7}=...=x_1>0$; then $x_{4i+2}=(4i+1)-x_1>4i\geq 0$; also $x_{4i+3}=(4i+2)-(4i+1)+x_1=1+x_1>0$; finally $x_{4i+4}=(4i+4)-(4i+2)+(4i+1)-x_1=4i+3-x_1>0$. Therefore, the answer is all $n$ not of the form $4k+2$ for some positive integer $k$.

This is essentially problem 8 from the shortlist of IMO 1992: https://artofproblemsolving.com/community/c6h220394p1222516