A paper equilateral triangle of side length $2$ on a table has vertices labeled $A,B,C.$ Let $M$ be the point on the sheet of paper halfway between $A$ and $C.$ Over time, point $M$ is lifted upwards, folding the triangle along segment $BM,$ while $A,B,$ and $C$ on the table. This continues until $A$ and $C$ touch. Find the maximum volume of tetrahedron $ABCM$ at any time during this process.
1. **Identify the key points and distances:**
   - The equilateral triangle \(ABC\) has side length \(2\).
   - Point \(M\) is the midpoint of \(AC\), so \(AM = MC = 1\).
   - The height of the equilateral triangle from \(B\) to \(AC\) is \(\sqrt{3}\).

2. **Determine the coordinates of the points:**
   - Place \(A\) at \((0, 0, 0)\), \(B\) at \((2, 0, 0)\), and \(C\) at \((1, \sqrt{3}, 0)\).
   - Point \(M\) is the midpoint of \(AC\), so \(M\) is at \((0.5, \frac{\sqrt{3}}{2}, 0)\).

3. **Folding the triangle:**
   - When \(M\) is lifted upwards, it moves along the \(z\)-axis while \(A\), \(B\), and \(C\) remain on the \(xy\)-plane.
   - Let the height of \(M\) above the \(xy\)-plane be \(h\).

4. **Volume of tetrahedron \(ABCM\):**
   - The base of the tetrahedron is the triangle \(ABC\) with area:
     \[
     \text{Area}_{ABC} = \frac{1}{2} \times 2 \times \sqrt{3} = \sqrt{3}
     \]
   - The volume \(V\) of the tetrahedron is given by:
     \[
     V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
     \]
     Here, the height is \(h\).

5. **Determine the maximum height \(h\):**
   - When \(A\) and \(C\) touch, \(M\) is directly above \(B\).
   - The distance \(AM\) and \(MC\) are both \(1\), and the height \(h\) forms a right triangle with \(AM\) and \(MC\).
   - Using the Pythagorean theorem in the right triangle \(AMB\):
     \[
     h = \sqrt{AM^2 - \left(\frac{AC}{2}\right)^2} = \sqrt{1^2 - \left(\frac{2}{2}\right)^2} = \sqrt{1 - 1} = 0
     \]
   - This implies that the maximum height \(h\) is \(\sqrt{3}\).

6. **Calculate the maximum volume:**
   - Substitute the maximum height \(h = \sqrt{3}\) into the volume formula:
     \[
     V = \frac{1}{3} \times \sqrt{3} \times \sqrt{3} = \frac{1}{3} \times 3 = 1
     \]

The final answer is \(\boxed{1}\)