Let $P_1$, $P_2$, $P_3$ be the vertices of an equilateral triangle with sides of length $1$ inscribed in a circle. Obtain points $Q_1$, $Q_2$, $Q_3$ by rotating the triangle $P_1P_2P_3$ so that $Q_i$ lies on the minor arc $P_iP_{i+1}$. Let $|P_1Q_1| = a$ and $|Q_1P_2| = b$. Applying Ptolemy's theorem to the cyclic quadrilateral $P_1Q_1P_2P_3$, we get $|Q_1P_3| = a + b$. Similarly $|Q_2P_1| = a + b = |Q_3P_2|$.

Applying the Cosine Law to triangle $P_1Q_1P_2$ and using that $\angle P_2Q_1P_1 = 180^\circ - \angle P_2P_3P_1 = 120^\circ$, we find that $a^2 + ab + b^2 = 1$. We wish to find solutions of this with $a$, $b$ rational and then scale up the diagram by an integer factor to obtain a hexagon in which all distances between vertices are integers. More specifically, we seek positive integers $m_1$, $m_2$, $n$ such that
$$\left(\frac{m_1}{n}\right)^2 + \frac{m_1m_2}{n^2} + \left(\frac{m_2}{n}\right)^2 = 1$$
i.e. $m_1^2 + m_1m_2 + m_2^2 = n^2$. By trial and error, we may find that $m_1 = 5$, $m_2 = 3$, $n = 7$ is such a solution. This leads to $a = 5/7$, $b = 3/7$ and then $a+b = 8/7$. We now need to scale up the original diagram by a factor of $7$.

The Pythagorean triple $(3, 4, 5)$ gives rise to two triples with a shared length: $(9, 12, 15)$ and $(12, 16, 20)$. The corresponding triangles $PAB$ and $QAC$ can be put together as in the diagram below:

Reflecting this diagram in the line $AQ$ and then in the perpendicular bisector of $PQ$, we obtain a hexagon $ABCDEF$ as shown in the next diagram.

The only distance that remains to be checked is $|BE|$, but this is the hypotenuse of a right angled triangle with sides $7$ and $24$, so $|BE| = 25$.

Let $P_x = (\cos(x), \sin(x))$ be the point on the unit circle given by the angle $x$ and recall that
$$\sin(\alpha \pm \beta) = \sin(\alpha) \cos(\beta) \pm \cos(\alpha) \sin(\beta) \\ \cos(\alpha \pm \beta) = \cos(\alpha) \cos(\beta) \mp \sin(\alpha) \sin(\beta).$$
This means that the points $P_{\alpha+\beta}$ and $P_{\alpha-\beta}$ have rational coordinates whenever this is the case for the two points $P_{\alpha}$ and $P_{\beta}$. Also, recall that the length of a chord $AB$ of the unit circle subtending a central angle of size $2\alpha$ is equal to $2\sin(\alpha)$. The consequence is that the chord $P_{2\alpha}P_{2\beta}$ has rational length whenever $P_{\alpha}$ and $P_{\beta}$ both have rational coordinates.

Hence, when we pick $n \ge 3$ points on the unit circle, all of the form $P_{2\alpha}$ where $P_{\alpha}$ has rational coordinates, we obtain a convex $n$-gon such that the distances between every pair of vertices is a rational number. Scaling up with the common denominator gives integer distances.

To construct explicit examples, recall that each Pythagorean triple $(a, b, c)$ gives rise to a point $(a/c, b/c)$ on the unit circle (and vice versa). For example, starting with
$$P_{\alpha} = (\cos(\alpha), \sin(\alpha)) = \left(\frac{4}{5}, \frac{3}{5}\right), \text{ we obtain}$$
$$P_{2\alpha} = (\cos(2\alpha), \sin(2\alpha)) = (\cos^2(\alpha) - \sin^2(\alpha), 2 \sin(\alpha) \cos(\alpha)) = \left(\frac{7}{25}, \frac{24}{25}\right).$$
When we use $D = (1,0)$, $E = P_{2\alpha}$, scale up by the factor $25/2$ and use symmetry we arrive at the example from Solution 2.