1. We start with the given equation in the problem:

$$\begin{array}{cccc} & G & O & G & O \\ + & D & I & D & I \\ \hline G & O & D & O & T \\ \end{array}$$

2. We need to determine the values of the digits $G, O, D, I,$ and $T$. Note that $G$ and $D$ cannot be $0$.

3. First, consider the units place:
$$O + I = T \quad \text{(1)}$$

4. Next, consider the tens place:
$$G + D + \text{carry from units place} = O \quad \text{(2)}$$

5. For the hundreds place:
$$O + I + \text{carry from tens place} = D \quad \text{(3)}$$

6. Finally, for the thousands place:
$$G + D + \text{carry from hundreds place} = G \quad \text{(4)}$$

7. From equation (4), since $G + D$ must result in $G$ with a carry, we have:
$$G + D = 10 + G \quad \Rightarrow \quad D = 10$$
However, $D$ must be a digit (0-9), so this is not possible. Therefore, there must be a carry from the previous place.

8. Reconsidering the thousands place with a carry:
$$G + D + 1 = G + 10 \quad \Rightarrow \quad D = 9$$

9. Substitute $D = 9$ into the tens place equation (2):
$$G + 9 + \text{carry from units place} = O$$
Since $G + 9$ must be at least 10 to have a carry, $G$ must be 1:
$$1 + 9 = 10 \quad \Rightarrow \quad \text{carry} = 1 \quad \Rightarrow \quad O = 0$$

10. Substitute $O = 0$ and $D = 9$ into the hundreds place equation (3):
$$0 + I + 1 = 9 \quad \Rightarrow \quad I = 8$$

11. Finally, substitute $O = 0$ and $I = 8$ into the units place equation (1):
$$0 + 8 = T \quad \Rightarrow \quad T = 8$$

12. Therefore, the digits are:
$$G = 1, \quad O = 0, \quad D = 9, \quad I = 8, \quad T = 8$$

13. The value of GODOT is:
$$GODOT = 10908$$

The final answer is $\boxed{10908}$.