Find all four-digit numbers $\overline{abcd}$ such that they are multiples of $3$ and that $\overline{ab}-\overline{cd}=11$.
($\overline{abcd}$ is a four-digit number; $\overline{ab}$ is a two digit-number as $\overline{cd}$ is).
1. We start with the given conditions:
   - $\overline{abcd}$ is a four-digit number.
   - $\overline{abcd}$ is a multiple of 3.
   - $\overline{ab} - \overline{cd} = 11$.

2. Let $\overline{abcd} = 1000a + 100b + 10c + d$, where $a, b, c, d$ are digits.

3. Let $\overline{ab} = 10a + b$ and $\overline{cd} = 10c + d$.

4. From the condition $\overline{ab} - \overline{cd} = 11$, we have:
   \[
   10a + b - (10c + d) = 11 \implies 10a + b - 10c - d = 11
   \]

5. We also know that $\overline{abcd}$ is a multiple of 3. A number is a multiple of 3 if the sum of its digits is a multiple of 3. Therefore:
   \[
   a + b + c + d \equiv 0 \pmod{3}
   \]

6. From the equation $10a + b - 10c - d = 11$, we can express $\overline{abcd}$ in terms of $\overline{cd}$:
   \[
   \overline{abcd} = 1000a + 100b + 10c + d
   \]
   Since $\overline{ab} = 10a + b$ and $\overline{cd} = 10c + d$, we can write:
   \[
   \overline{abcd} = 100 \cdot (10a + b) + \overline{cd} = 100 \cdot (\overline{ab}) + \overline{cd}
   \]
   Using $\overline{ab} = \overline{cd} + 11$, we get:
   \[
   \overline{abcd} = 100 \cdot (\overline{cd} + 11) + \overline{cd} = 100 \overline{cd} + 1100 + \overline{cd} = 1100 + 101 \overline{cd}
   \]

7. Since $\overline{abcd}$ is a multiple of 3, we need $1100 + 101 \overline{cd}$ to be a multiple of 3. We check the remainders modulo 3:
   \[
   1100 \equiv 2 \pmod{3}
   \]
   \[
   101 \equiv 2 \pmod{3}
   \]
   Therefore:
   \[
   1100 + 101 \overline{cd} \equiv 2 + 2 \overline{cd} \equiv 0 \pmod{3}
   \]
   Simplifying, we get:
   \[
   2 + 2 \overline{cd} \equiv 0 \pmod{3} \implies 2 \overline{cd} \equiv 1 \pmod{3}
   \]
   Multiplying both sides by the modular inverse of 2 modulo 3, which is 2, we get:
   \[
   \overline{cd} \equiv 2 \pmod{3}
   \]
   Therefore, $\overline{cd}$ must be of the form $3n + 2$.

8. We need to find all possible values of $\overline{cd}$ such that $\overline{cd}$ is a two-digit number and $\overline{cd} \equiv 2 \pmod{3}$. The possible values are:
   \[
   \overline{cd} \in \{2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98\}
   \]

9. For each value of $\overline{cd}$, we calculate $\overline{ab} = \overline{cd} + 11$ and check if the resulting four-digit number $\overline{abcd}$ is a multiple of 3.

10. For example, if $\overline{cd} = 2$, then $\overline{ab} = 2 + 11 = 13$. The number $\overline{abcd} = 1302$:
    \[
    1 + 3 + 0 + 2 = 6 \equiv 0 \pmod{3}
    \]
    So, $1302$ is a valid number.

11. Repeating this process for all values of $\overline{cd}$, we find the valid four-digit numbers.

The final answer is $\boxed{1302, 1605, 1908, 2211, 2514, 2817, 3120, 3423, 3726, 4029, 4332, 4635, 4938, 5241, 5544, 5847, 6150, 6453, 6756, 7059, 7362, 7665, 7968, 8271, 8574, 8877, 9180, 9483, 9786, 10089, 10392, 10695, 10998}$