Variable points $M$ and $N$ are considered on the arms $\left[ OX \right.$ and $\left[ OY \right.$ , respectively, of an angle $XOY$ so that $\left| OM \right|+\left| ON \right|$ is constant. Determine the locus of the midpoint of $\left[ MN \right]$.
1. **Define the problem and setup:**
   We are given an angle \(XOY\) with variable points \(M\) and \(N\) on the arms \([OX]\) and \([OY]\) respectively, such that \(|OM| + |ON| = k\), where \(k\) is a constant. We need to determine the locus of the midpoint of the segment \([MN]\).

2. **Introduce sample points and properties:**
   Let \(M'\) and \(N'\) be points on \([OX]\) and \([OY]\) respectively, such that \(OM' = ON' = \frac{k}{2}\). Since \(k\) is constant, the angle \(\angle M'ON'\) is also constant, making the segment \([M'N']\) constant in length.

3. **Construct auxiliary lines and points:**
   Draw a line through \(N\) parallel to \(OM\). Let this line intersect \(M'N'\) at point \(Q\). Let \(MN\) intersect \(M'N'\) at point \(P\).

4. **Establish relationships and properties:**
   Since \(OM + ON = OM' + ON' = k\), and given that \(OM' = ON' = \frac{k}{2}\), it follows that \(MM' = NN'\). Also, since \(\angle OM'N' = \angle NQN'\), we have \(NQ = NN' = MM'\).

5. **Midpoint property:**
   Since \(MP = PN\), the midpoint of the segment \(MN\) lies on the segment \(M'N'\).

6. **Prove the inverse:**
   To show that any point \(P\) on \([M'N']\) can be the midpoint of some segment \(MN\) with \(OM + ON = k\), consider the reflection of \(O\) over \(P\), denoted as \(O'\). Draw parallels through \(O'\) to \(ON'\) and \(OM'\), meeting \(OM'\) and \(ON'\) at points \(M\) and \(N\) respectively.

7. **Verify the construction:**
   Since \(ONO'M\) forms a parallelogram, \(P\) is the midpoint of \(MN\). To confirm \(OM + ON = k\), let \(O'N\) intersect \(M'N'\) at \(Q\). Given that \(\triangle OM'N'\) is isosceles, \(\triangle NQN'\) is also isosceles, implying \(NQ = NN'\). Since \(MM' = NQ\), it follows that \(OM + ON = OM' + ON' = k\).

8. **Conclusion:**
   The midpoint of the segment \(MN\) always lies on the segment \(M'N'\).

\(\blacksquare\)

The final answer is the locus of the midpoint of \( \boxed{ [MN] } \) is the segment \([M'N']\).