Solution

Let us write down the statics equation relative to a point that will be stationary at the beginning of the movement. Conventionally, the distance $x$ from the point of application of force $F$. On different sides of this point, there will be different masses of the block.
$$\mu m_1g\frac{x}{2}+\mu m_2\frac{L-x}{2}=Fx$$
From the homogeneity of the rod it is clear that:
$$m_1=\frac{m}{L}x$$
$$m_2=\frac{m}{L}(L-x)$$
Let's substitute:
$$\mu mg\frac{x^2}{2L}+\mu mg\frac{(L-x)^2}{2L}=Fx$$
Let's open the brackets by dividing both sides of the equation by $x$:
$$\mu mg(\frac{x}{L}+\frac{L}{2x}-1)=F$$
Let's take the derivative of the force along the shoulder of the applied force, because we need a minimum:
$$\frac{d}{dx}F=\frac{1}{L}-\frac{L}{2x^2}=0$$
We express the shoulder in units of rod length:
$$x=\frac{L}{\sqrt{2}}$$
Let's substitute the forces into the equation:
$$\mu mg(\frac{\frac{L}{\sqrt{2}}}{L}+\frac{L}{2\frac{L}{\sqrt{2}}}-1)=F$$
Let's reduce the length and rationalize the fractions:
$$\mu mg(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-1)=F$$
From here we get the answer:
$$\boxed{F=\mu mg(\sqrt{2}-1)}$$