Let $ABCDE$ be a convex pentagon in which $\angle{A}=\angle{B}=\angle{C}=\angle{D}=120^{\circ}$ and the side lengths are five [i]consecutive integers[/i] in some order. Find all possible values of $AB+BC+CD$.
Given a convex pentagon \(ABCDE\) with \(\angle A = \angle B = \angle C = \angle D = 120^\circ\) and side lengths that are five consecutive integers in some order, we need to find all possible values of \(AB + BC + CD\).

1. **Assigning Side Lengths:**
   Let the side lengths be \(a-2, a-1, a, a+1, a+2\) in some order, where \(a \in \mathbb{N}\).

2. **Extending Lines to Form Equilateral Triangles:**
   Extend \(EA\), \(ED\), and \(BC\). Since \(\angle E = 60^\circ\), we form equilateral triangles. This gives us the conditions:
   \[
   EA = BC + CD \quad \text{and} \quad DE = BA + BC
   \]
   We also note that \(EA > BC, CD\) and \(DE > BA, BC\).

3. **Case Analysis:**
   We assume \(EA > DE\) and analyze the possible cases. We need to check the validity of each case by ensuring the conditions \(EA = BC + CD\) and \(DE = BA + BC\) are satisfied.

   - **Case 1:** \(EA > DE > BC > BA > CD\)
     \[
     (EA, DE, BC, BA, CD) = (a+2, a+1, a, a-1, a-2)
     \]
     From \(EA = BC + CD\):
     \[
     a+2 = a + (a-2) \implies a = 4
     \]
     Check \(DE = BA + BC\):
     \[
     a+1 = (a-1) + a \implies 5 = 3 + 4 \quad \text{(valid)}
     \]

   - **Case 2:** \(EA > DE > BA > BC > CD\)
     \[
     (EA, DE, BA, BC, CD) = (a+2, a+1, a-1, a, a-2)
     \]
     From \(EA = BC + CD\):
     \[
     a+2 = a + (a-2) \implies a = 4
     \]
     Check \(DE = BA + BC\):
     \[
     a+1 = (a-1) + a \implies 5 = 3 + 4 \quad \text{(valid)}
     \]

   - **Case 3:** \(EA > DE > CD > BC > BA\)
     \[
     (EA, DE, CD, BC, BA) = (a+2, a+1, a-2, a, a-1)
     \]
     From \(EA = BC + CD\):
     \[
     a+2 = a + (a-2) \implies a = 4
     \]
     Check \(DE = BA + BC\):
     \[
     a+1 = (a-1) + a \implies 5 = 3 + 4 \quad \text{(valid)}
     \]

4. **Swapping \(DE\) and \(EA\):**
   We now assume \(DE > EA\) and swap the values \(DE \leftrightarrow EA\) and \(BA \leftrightarrow CD\) in the valid cases above.

   - **Case 4:** \(DE > EA > BC > BA > CD\)
     \[
     (DE, EA, BC, BA, CD) = (a+2, a+1, a, a-1, a-2)
     \]
     From \(DE = BA + BC\):
     \[
     a+2 = (a-1) + a \implies a = 4
     \]
     Check \(EA = BC + CD\):
     \[
     a+1 = a + (a-2) \implies 5 = 4 + 2 \quad \text{(valid)}
     \]

   - **Case 5:** \(DE > EA > BA > BC > CD\)
     \[
     (DE, EA, BA, BC, CD) = (a+2, a+1, a-1, a, a-2)
     \]
     From \(DE = BA + BC\):
     \[
     a+2 = (a-1) + a \implies a = 4
     \]
     Check \(EA = BC + CD\):
     \[
     a+1 = a + (a-2) \implies 5 = 4 + 2 \quad \text{(valid)}
     \]

   - **Case 6:** \(DE > EA > CD > BC > BA\)
     \[
     (DE, EA, CD, BC, BA) = (a+2, a+1, a-2, a, a-1)
     \]
     From \(DE = BA + BC\):
     \[
     a+2 = (a-1) + a \implies a = 4
     \]
     Check \(EA = BC + CD\):
     \[
     a+1 = a + (a-2) \implies 5 = 4 + 2 \quad \text{(valid)}
     \]

5. **Summarizing Possible Values:**
   The valid cases give us the following possible values for \(AB + BC + CD\):
   \[
   AB + BC + CD \in \{6, 9, 7\}
   \]

The final answer is \( \boxed{ \{6, 9, 7\} } \).