Solution

Let's renumber the weights as shown in the figure, and direct the $X$ axis to the right.

It is clear that then none of the weights can have negative acceleration.

Let's prove that weights $3$ and $4$ move as a single whole. To do this, let's assume the opposite: let weight $3$ slide on weight $4$. Then a friction force arises between them

$$F_{fr}= \mu mg$$

and elastic force arises in the thread

$$T> \mu mg$$

In this case, the acceleration of load $2$ would be directed to the left, which is impossible. Therefore, the accelerations of loads $2$, $3$, and $4$ are the same.

Let us denote the acceleration of these loads as $\bar{a}_{1}=\bar{a}_{2}$, and the acceleration of load $1$ as $\bar{a}_{1}$.

Now consider two cases.

Case 1. Let loads $1$ and $2$ be at relative rest and $\bar{a}_{1}=\bar{a}_{2}$.

Let us denote the modulus of the static friction force between them as $F_1$, the modulus of the friction force between loads $3$ and $4$ as $F_2$, and the modulus of the elastic force of the thread as $T$ .

Then: for cargo $1$

$$F-F_{1}=Ma_{1}$$

for cargo $2$

$$F_{1}-T=ma_{2}$$

for cargo $3$

$$T-F_{2}=ma_{2}$$

for cargo $4$

$$F_{2}=Ma_{2}$$

Solving this system of equations, we obtain:

$$F_{1}=\frac{2m+M}{2(M+m)}F,a_{1}=a_{2}=\frac{F}{2(M+m)}$$

The same result can be obtained in another way. Since the friction between all surfaces is static friction, the system of loads moves as one body with mass $M=2(M+m)$.

Therefore

$$\bar{F}=m \bar{a}_{1},\bar{a}_{1}=\bar{a}_{2}= \frac{\bar{F}}{2(M+m)}$$

Case $2$.

Let load $2$ slide on load $1$. Then the friction force acts on load $1$

$$F_{fr}^{\prime}= \mu mg$$

and this load receives acceleration $a_{1}=\frac{F- \mu mg}{m}$.

The system of loads $2$, $3$ and $4$ moves as one body, the mass of which is $M_{0}=2m+M$ with acceleration

$$a_{2}=\frac{\mu mg}{2m+M}$$

The first case is realized if

$$F \geq \frac{2 \mu m (m+M)g}{2m+M}$$