A right-triangular wooden block of mass $M$ is at rest on a table, as shown in figure. Two smaller wooden cubes, both with mass $m$, initially rest on the two sides of the larger block. As all contact surfaces are frictionless, the smaller cubes start sliding down the larger block while the block remains at rest. What is the normal force from the system to the table?

$\textbf{(A) } 2mg\\
\textbf{(B) }  2mg + Mg\\
\textbf{(C) } mg + Mg\\
\textbf{(D) } Mg + mg( \sin \alpha + \sin \beta)\\
\textbf{(E) } Mg + mg( \cos \alpha + \cos \beta)$
1. **Identify the forces acting on the system:**
   - The larger block of mass \( M \) is at rest on the table.
   - Two smaller cubes, each of mass \( m \), are sliding down the sides of the larger block.
   - All contact surfaces are frictionless.

2. **Determine the normal forces exerted by the smaller cubes on the larger block:**
   - For the cube on the left side, the normal force exerted by the larger block on the cube is \( N_1 = mg \cos \alpha \).
   - For the cube on the right side, the normal force exerted by the larger block on the cube is \( N_2 = mg \cos \beta \).

3. **Calculate the vertical components of these normal forces:**
   - The vertical component of the normal force from the left cube is \( N_{1y} = N_1 \cos \alpha = mg \cos^2 \alpha \).
   - The vertical component of the normal force from the right cube is \( N_{2y} = N_2 \cos \beta = mg \cos^2 \beta \).

4. **Sum the vertical forces acting on the system:**
   - The total weight of the system is the sum of the weights of the larger block and the two smaller cubes: \( Mg + 2mg \).
   - The normal force from the table must balance this total weight.

5. **Calculate the normal force from the table:**
   - The normal force from the table, \( N \), must support the weight of the larger block and the vertical components of the normal forces from the smaller cubes.
   - Therefore, \( N = Mg + mg \cos^2 \alpha + mg \cos^2 \beta \).

6. **Simplify the expression:**
   - Since the total weight of the system is \( Mg + 2mg \), and the vertical components of the normal forces from the smaller cubes are already included in the weight calculation, the normal force from the table is:
   \[
   N = Mg + mg \cos^2 \alpha + mg \cos^2 \beta
   \]

7. **Compare with the given options:**
   - The correct option is \( Mg + mg(\cos^2 \alpha + \cos^2 \beta) \).

The final answer is \( \boxed{Mg + mg(\cos^2 \alpha + \cos^2 \beta)} \).