1. Let the vertices of the triangle be $(0, 0)$, $(a, b)$, and $(c, d)$, where $a, b, c, d$ are integers.
2. The lengths of the sides of the triangle are given by:
$$\text{Length}_1 = \sqrt{a^2 + b^2}$$
$$\text{Length}_2 = \sqrt{c^2 + d^2}$$
$$\text{Length}_3 = \sqrt{(a-c)^2 + (b-d)^2}$$
3. For all sides to be odd, the squares of the lengths must be congruent to 1 modulo 4. Therefore, we have:
$$a^2 + b^2 \equiv 1 \pmod{4}$$
$$c^2 + d^2 \equiv 1 \pmod{4}$$
$$(a-c)^2 + (b-d)^2 \equiv 1 \pmod{4}$$
4. Consider the possible values of $a^2$ and $b^2$ modulo 4. Since $a$ and $b$ are integers, $a^2$ and $b^2$ can be 0 or 1 modulo 4:
$$a^2 \equiv 0 \text{ or } 1 \pmod{4}$$
$$b^2 \equiv 0 \text{ or } 1 \pmod{4}$$
5. To satisfy $a^2 + b^2 \equiv 1 \pmod{4}$, the possible combinations are:
$$(a^2, b^2) \equiv (0, 1) \text{ or } (1, 0) \pmod{4}$$
6. Similarly, for $c^2 + d^2 \equiv 1 \pmod{4}$, the possible combinations are:
$$(c^2, d^2) \equiv (0, 1) \text{ or } (1, 0) \pmod{4}$$
7. Now consider the expression $(a-c)^2 + (b-d)^2$:
$$(a-c)^2 + (b-d)^2 = a^2 + c^2 - 2ac + b^2 + d^2 - 2bd$$
$$\equiv (a^2 + b^2) + (c^2 + d^2) - 2(ac + bd) \pmod{4}$$
8. Since $a^2 + b^2 \equiv 1 \pmod{4}$ and $c^2 + d^2 \equiv 1 \pmod{4}$, we have:
$$(a^2 + b^2) + (c^2 + d^2) \equiv 1 + 1 \equiv 2 \pmod{4}$$
9. Therefore:
$$(a-c)^2 + (b-d)^2 \equiv 2 - 2(ac + bd) \pmod{4}$$
10. For $(a-c)^2 + (b-d)^2$ to be congruent to 1 modulo 4, we need:
$$2 - 2(ac + bd) \equiv 1 \pmod{4}$$
11. This simplifies to:
$$2 \equiv 1 + 2(ac + bd) \pmod{4}$$
$$1 \equiv 2(ac + bd) \pmod{4}$$
12. Since $2(ac + bd)$ is always even, it can only be 0 or 2 modulo 4. Therefore, $2(ac + bd) \equiv 0 \text{ or } 2 \pmod{4}$, which means:
$$1 \equiv 0 \text{ or } 2 \pmod{4}$$
13. This is a contradiction because 1 is not congruent to 0 or 2 modulo 4. Therefore, it is impossible to find such a triangle with all integer coordinates and all sides of odd length.