First, no even integer $n$ can be among the integers sought. Indeed, the inscribed circle of an equilateral triangle with side 1 has an area $\mathcal{A}=\sqrt{3} / 4$ and a radius $r=2 \mathcal{A} / 3=\sqrt{3} / 6$. Consequently, the almond shape below, drawn in the case where $n=6$, and itself contained within a band of width $r$, could not contain such an equilateral triangle.
!

Conversely, let $n \geqslant 3$ be an integer for which there exists a convex polygon $\mathcal{P}=A_{0} A_{1} \ldots A_{n-1}$, with side 1, as described in the statement, and which does not contain any equilateral triangle of side 1. We assume that its vertices have been listed in a clockwise direction. Inspired by the above construction, we denote by $d$ the maximum distance between two points of $\mathcal{P}$, and, without loss of generality, we assume that there exists a point $A_{\ell}$, with $1 \leqslant \ell \leqslant k$, located at a distance $d$ from $A_{0}$.
If $d=1$, the polygon $\mathcal{P}$ is contained in the gray area represented below, which consists of the intersection of two semicircles of radius 1. Since $A_{2}$ is at a distance 1 from $A_{1}$, it lies on the arc $\widehat{B A_{0}}$ and, similarly, $A_{n-1}$ lies on the arc $\widehat{A_{1} B}$. Since $\mathcal{P}$ is convex, we deduce that $A_{2}=B=A_{n-1}$, so that $\mathcal{P}$ is an equilateral triangle, which of course contains itself.
!

We therefore know that $d>1$, so that $A_{0} A_{\ell}$ is a diagonal of $\mathcal{P}$. If $\widehat{A_{\ell} A_{0} A_{1}} \geqslant 60^{\circ}$, and since $A_{1} A_{\ell} \leqslant d=A_{0} A_{\ell}$, we know that $\widehat{A_{0} A_{1} A_{\ell}} \geqslant \widehat{A_{\ell} A_{0} A_{1}} \geqslant 60^{\circ}$, so that the triangle $A_{0} A_{1} A_{\ell}$ itself contains an equilateral triangle of side 1, as illustrated below.
!

Thus, $\widehat{A_{\ell} A_{0} A_{1}}<60^{\circ}$ and, similarly, $\widehat{A_{\ell-1} A_{\ell}} A_{0}<60^{\circ}$. We then draw the isosceles trapezoid $\mathcal{T}=A_{0} B_{0} B_{\ell} A_{\ell}$ represented below in the case where $\ell=4$, with angles $60^{\circ}$ at $A_{0}$ and $A_{\ell}$ and sides $A_{0} B_{0}=A_{\ell} B_{\ell}=1$. If there exists a point $A_{i}$ (with $1 \leqslant i \leqslant \ell-1$) located outside $\mathcal{T}$, the polygon $\mathcal{P}$ again contains an equilateral triangle of side 1, as illustrated below, in white on a gray background.
!

This case does not occur either, so the polygon $A_{0} A_{1} \ldots A_{\ell}$ is strictly included in the polygon $\mathcal{T}$. Its perimeter $\ell+d$ is therefore strictly less than that of $\mathcal{T}$, i.e., $2 d+1$, so $\ell-1<d$, and the triangle inequality also indicates that $d \leqslant \ell$, which means that $\ell=\lceil d\rceil$. By considering the polygon $A_{0} A_{n} A_{n-1} \ldots A_{\ell}$, we similarly conclude that $n-\ell=\lceil d\rceil$, so that $n=2\lceil d\rceil$ is necessarily even.
Thus, the integers sought are the odd integers.
Remark: We stated above that when a convex polygon $\mathcal{P}$ is strictly included in a polygon $\mathcal{Q}$, the perimeter of $\mathcal{P}$ is strictly less than that of $\mathcal{Q}$. This "classical" statement can be demonstrated as follows.
First, let $\mathcal{C}$ be the convex hull of $\mathcal{Q}$. By construction, its perimeter is less than or equal to that of $\mathcal{Q}$, and we can therefore assume without loss of generality that $\mathcal{Q}$ is convex.
In the following, we assume $\mathcal{Q}$ is fixed, then we denote by $\mathrm{p}(\mathcal{P})$ the perimeter of a polygon $\mathcal{P}$ and $\mathrm{c}(\mathcal{P})$ the number of sides of $\mathcal{P}$ that are not included in sides of $\mathcal{Q}$. We then prove by induction on $c(\mathcal{P})$ that $p(\mathcal{P}) \leqslant p(\mathcal{Q})$.
Given a polygon $\mathcal{P}$ strictly included in $\mathcal{Q}$, we know that $c(\mathcal{P}) \geqslant 1$, and we consider a side $[B C]$ of $\mathcal{P}$ that is not included in a side of $\mathcal{Q}$. Since $\mathcal{Q}$ is convex, this side is strictly included in $\mathcal{Q}$. Then, as illustrated below, we extend the rays $A B)$ and $[D C)$ to infinity, or until they meet. We thus define a finite or infinite region, which we denote by $\mathcal{Z}$, and which may be bounded or unbounded.
![

Case 1
!

Case 2
!

Case 3

We then integrate into the polygon $\mathcal{P}$ the portion of the polygon $\mathcal{Q}$ included in the region $\mathcal{Z}$, which we have represented in blue. We thus form a new convex polygon $\mathcal{P}^{\prime}$. Since the side $[B C]$ of $\mathcal{P}$ has disappeared, we know that $\mathrm{c}\left(\mathcal{P}^{\prime}\right)<\mathrm{c}(\mathcal{P})$. The induction hypothesis and the triangle inequality then indicate that $\mathrm{p}(\mathcal{P})<\mathrm{p}\left(\mathcal{P}^{\prime}\right) \leqslant \mathrm{p}(\mathcal{Q})$, which concludes.

Comment from the graders: This problem was extremely difficult, to the point that only one student scored more than two points (and solved it). Many students had the excellent idea of using an elongated almond to handle the case $n$ even. The main idea, difficult to find, was then to continue with an analogous construction by considering the elongation of a polygon, i.e., the greatest distance between two vertices. It is unfortunate, however, that several students claimed to have "magically" solved the case $n$ odd: such claims will obviously not help to score points, and their only possible effect would be to annoy the grader (which did not happen here).