In a triangle $ABC$, $M$ is the midpoint of $BC$. Given that $\angle MAC=15^{\circ}$, find the maximum possible value of $\angle ABC$.
1. **Draw the circumcircle of $\Delta ACM$ with center $O$.** 
   - Since $M$ is the midpoint of $BC$, $M$ is equidistant from $B$ and $C$.
   - $\angle MAC = 15^\circ$ implies that $\angle MCA = \angle MBC = 15^\circ$ because $M$ is the midpoint and $\angle MAC$ is subtended by the arc $MC$.

2. **Consider the point $X$ on the circumcircle such that $\angle CXM = 15^\circ$.**
   - By the property of the circle, any point $X$ on the circumcircle of $\Delta ACM$ will satisfy $\angle CXM = 15^\circ$.

3. **Determine the maximum value of $\angle ABC$.**
   - The maximum value of $\angle ABC$ is obtained when $BX$ is tangent to the circumcircle at $X$.
   - By the tangent-secant theorem (or power of a point), $BX^2 = BM \cdot BC$.

4. **Use the Law of Sines in $\Delta BXM$ and $\Delta CXM$.**
   - In $\Delta BXM$:
     \[
     \frac{BX}{\sin \angle BXM} = \frac{BM}{\sin \angle BMX}
     \]
   - In $\Delta CXM$:
     \[
     \frac{CX}{\sin \angle CXM} = \frac{CM}{\sin \angle CMX}
     \]

5. **Relate the sides and angles using the given conditions.**
   - Since $M$ is the midpoint of $BC$, $BM = CM = \frac{BC}{2}$.
   - $\angle CXM = 15^\circ$ and $\angle CMX = 180^\circ - \angle CXM - \angle XCM = 180^\circ - 15^\circ - \angle XCM$.

6. **Simplify the expressions using the Law of Sines.**
   - From the Law of Sines in $\Delta BXM$:
     \[
     \frac{BX}{\sin \angle BXM} = \frac{BM}{\sin \angle BMX}
     \]
     Since $BX$ is tangent to the circle at $X$, $\angle BXM = 90^\circ$ and $\sin \angle BXM = 1$.
     \[
     BX = BM \cdot \sin \angle BMX
     \]

7. **Calculate the maximum value of $\angle ABC$.**
   - The maximum value of $\angle ABC$ occurs when $\angle BXM = 90^\circ$.
   - Therefore, $\angle ABC = 2 \times 15^\circ = 30^\circ$.

The final answer is $\boxed{30^\circ}$.