To solve this problem, we can use the equations of motion under uniform acceleration, separately considering the horizontal and vertical motions because the two are independent of each other.

First, for the body of mass $M$ thrown horizontally with velocity $v$ from a height $H$, let's analyze its motion:

Horizontal Motion:

The horizontal distance (range) $x$ covered by the object is given by $x = v \cdot t$, where $t$ is the time taken to hit the ground.

Vertical Motion:

The time $t$ it takes for the object to hit the ground can be found using the equation of motion under gravity, $H = \frac{1}{2} g t^2$, where $g$ is the acceleration due to gravity.

For the first body:

Given $x = 100$ m and using the equation for the vertical motion to find $t$, we have:

$H = \frac{1}{2} g t^2$

Solving for $t$, we get:

$t = \sqrt{\frac{2H}{g}}$

The horizontal motion gives:

$x = v \cdot t \Rightarrow 100 = v \cdot \sqrt{\frac{2H}{g}}$

Now, considering the second body of mass $2M$ thrown at velocity $\frac{v}{2}$ from height $4H$:

For the second body:

The time $t'$ it takes for the second body to hit the ground from height $4H$ can be found by:

$4H = \frac{1}{2} g t'^2$

Solving for $t'$, we get:

$t' = \sqrt{\frac{2 \cdot 4H}{g}} = 2 \sqrt{\frac{2H}{g}}$

This is twice the time $t$ found for the first body.

The horizontal distance $x'$ covered by the second body is:

$x' = \left(\frac{v}{2}\right) \cdot t'$

Now, substituting the value of $t'$ found above, we get:

$x' = \left(\frac{v}{2}\right) \cdot 2\sqrt{\frac{2H}{g}} = v \cdot \sqrt{\frac{2H}{g}}$

But we previously found that $v \cdot \sqrt{\frac{2H}{g}} = 100$, so:

$x' = 100 \text{ m}$

Therefore, a body of mass $2M$ thrown horizontally with velocity $v/2$ from the top of a tower of height $4H$ will touch the ground at a distance of 100 meters from the foot of the tower.