IMO Shortlist, 2021

Answer: All odd $n \geqslant 3$.

First we show that for every even $n \geqslant 4$ there exists a polygon violating the required statement. Consider a regular $k$-gon $A_{0} A_{1}, \ldots A_{k-1}$ with side length $1$. Let $B_{1}, B_{2}, \ldots, B_{n / 2-1}$ be the points symmetric to $A_{1}, A_{2}, \ldots, A_{n / 2-1}$ with respect to the line $A_{0} A_{n / 2}$. Then $P= A_{0} A_{1} A_{2} \ldots A_{n / 2-1} A_{n / 2} B_{n / 2-1} B_{n / 2-2} \ldots B_{2} B_{1}$ is a convex $n$-gon whose sides all have length $1$. If $k$ is big enough, $P$ is contained in a strip of width $1 / 2$, which clearly does not contain any equilateral triangle of side length $1$.

Assume now that $n=2k+1$. As the case $k=1$ is trivially true, we assume $k \geqslant 2$ henceforth. Consider a convex $(2k+1)$-gon $P$ whose sides all have length $1$. Let $d$ be its longest diagonal. The endpoints of $d$ split the perimeter of $P$ into two polylines, one of which has length at least $k+1$. Hence we can label the vertices of $P$ so that $P=A_{0} A_{1} \ldots A_{2k}$ and $d=A_{0} A_{\ell}$ with $\ell \geqslant k+1$. We will show that, in fact, the polygon $A_{0} A_{1} \ldots A_{\ell}$ contains an equilateral triangle of side length $1$.

Suppose that $\angle A_{\ell} A_{0} A_{1} \geqslant 60^{\circ}$. Since $d$ is the longest diagonal, we have $A_{1} A_{\ell} \leqslant A_{0} A_{\ell}$, so $\angle A_{0} A_{1} A_{\ell} \geqslant \angle A_{\ell} A_{0} A_{1} \geqslant 60^{\circ}$. It follows that there exists a point $X$ inside the triangle $A_{0} A_{1} A_{\ell}$ such that the triangle $A_{0} A_{1} X$ is equilateral, and this triangle is contained in $P$. Similar arguments apply if $\angle A_{\ell-1} A_{\ell} A_{0} \geqslant 60^{\circ}$.

From now on, assume $\angle A_{\ell} A_{0} A_{1}<60^{\circ}$ and $A_{\ell-1} A_{\ell} A_{0}<60^{\circ}$.
Consider an isosceles trapezoid $A_{0} Y Z A_{\ell}$ such that $A_{0} A_{\ell} \parallel Y Z$, $A_{0} Y=Z A_{\ell}=1$, and $\angle A_{\ell} A_{0} Y=\angle Z A_{\ell} A_{0}=60^{\circ}$. Suppose that $A_{0} A_{1} \ldots A_{\ell}$ is contained in $A_{0} Y Z A_{\ell}$. Note that the perimeter of $A_{0} A_{1} \ldots A_{\ell}$ equals $\ell+A_{0} A_{\ell}$ and the perimeter of $A_{0} Y Z A_{\ell}$ equals $2 A_{0} A_{\ell}+1$.

Recall a well-known fact stating that if a convex polygon $P_{1}$ is contained in a convex polygon $P_{2}$, then the perimeter of $P_{1}$ is at most the perimeter of $P_{2}$. Hence we obtain
$$\ell+A_{0} A_{\ell} \leqslant 2 A_{0} A_{\ell}+1, \quad \text{ i.e. } \quad \ell-1 \leqslant A_{0} A_{\ell} .$$
On the other hand, the triangle inequality yields
$$A_{0} A_{\ell}<A_{\ell} A_{\ell+1}+A_{\ell+1} A_{\ell+2}+\ldots+A_{2k} A_{0}=2k+1-\ell \leqslant \ell-1,$$
which gives a contradiction.

Therefore, there exists a vertex $A_{m}$ of $A_{0} A_{1} \ldots A_{\ell}$ which lies outside $A_{0} Y Z A_{\ell}$. Since
$$\begin{equation*} \angle A_{\ell} A_{0} A_{1}<60^{\circ}=\angle A_{\ell} A_{0} Y \quad \text{ and } \quad A_{\ell-1} A_{\ell} A_{0}<60^{\circ}=\angle Z A_{\ell} A_{0}, \tag{1} \end{equation*}$$
the distance between $A_{m}$ and $A_{0} A_{\ell}$ is at least $\sqrt{3} / 2$.
Let $P$ be the projection of $A_{m}$ to $A_{0} A_{\ell}$. Then $P A_{m} \geqslant \sqrt{3} / 2$, and by (1) we have $A_{0} P>1 / 2$ and $P A_{\ell}>1 / 2$. Choose points $Q \in A_{0} P$, $R \in P A_{\ell}$, and $S \in P A_{m}$ such that $P Q=P R=1 / 2$ and $P S=\sqrt{3} / 2$. Then $Q R S$ is an equilateral triangle of side length $1$ contained in $A_{0} A_{1} \ldots A_{\ell}$.