Given that $A,B,$ and $C$ are noncollinear points in the plane with integer coordinates such that the distances $AB,AC,$ and $BC$ are integers, what is the smallest possible value of $AB?$
To find the smallest possible value of \( AB \) given that \( A, B, \) and \( C \) are noncollinear points with integer coordinates and the distances \( AB, AC, \) and \( BC \) are integers, we will analyze the problem step-by-step.

1. **Upper Bound Consideration**:
   Notice that a triangle with side lengths 3, 4, and 5 forms a right triangle with integer coordinates. This gives us an upper bound of 3 for \( AB \).

2. **Parity Analysis**:
   We use the fact that for any integer \( a \), \( a \equiv a^2 \pmod{2} \). This implies that \( a \) and \( a^2 \) have the same parity (both are either even or odd).

   Applying this, we get:
   \[
   AB + BC + AC \equiv AB^2 + BC^2 + AC^2 \pmod{2}
   \]

3. **Coordinate Assignment**:
   Assign coordinates to points \( A, B, \) and \( C \) as \( (a_1, a_2), (b_1, b_2), (c_1, c_2) \) respectively. Substituting these into the equation, we have:
   \[
   AB^2 = (a_1 - b_1)^2 + (a_2 - b_2)^2
   \]
   \[
   BC^2 = (b_1 - c_1)^2 + (b_2 - c_2)^2
   \]
   \[
   AC^2 = (a_1 - c_1)^2 + (a_2 - c_2)^2
   \]

4. **Simplifying the Parity Equation**:
   \[
   AB + BC + AC \equiv (a_1 - b_1)^2 + (a_2 - b_2)^2 + (b_1 - c_1)^2 + (b_2 - c_2)^2 + (a_1 - c_1)^2 + (a_2 - c_2)^2 \pmod{2}
   \]
   \[
   AB + BC + AC \equiv (a_1 - b_1) + (c_1 - a_1) + (b_1 - c_1) + (a_2 - b_2) + (c_2 - a_2) + (b_2 - c_2) \pmod{2}
   \]
   \[
   AB + BC + AC \equiv 0 \pmod{2}
   \]

5. **Case Analysis**:
   - **Case 1: \( AB = 1 \)**
     \[
     |AC - BC| < AB \implies |AC - BC| < 1 \implies |AC - BC| = 0 \implies AC = BC
     \]
     Then,
     \[
     AB + BC + AC = 1 + 2(AC) \equiv 0 \pmod{2}
     \]
     This is not possible since \( 1 + 2(AC) \) is odd.
     \[
     \implies AB \neq 1
     \]

   - **Case 2: \( AB = 2 \)**
     \[
     |AC - BC| < AB \implies |AC - BC| < 2 \implies |AC - BC| = 0 \text{ or } 1
     \]
     If \( |AC - BC| = 0 \), then \( AC = BC \), which is not possible as shown in Case 1.
     If \( |AC - BC| = 1 \), then:
     \[
     AC = BC + 1
     \]
     \[
     AB + BC + AC = 2 + BC + (BC + 1) = 3 + 2BC
     \]
     \[
     3 + 2BC \equiv 0 \pmod{2}
     \]
     This is not possible since \( 3 + 2BC \) is odd.
     \[
     \implies AB \neq 2
     \]

6. **Conclusion**:
   Since \( AB \neq 1 \) and \( AB \neq 2 \), the smallest possible value for \( AB \) is 3.

The final answer is \( \boxed{3} \).