A small block of mass m is placed on a bigger block of mass M that...

Let the upward direction be taken as positive.

When the system is displaced by a distance $x$ above its equilibrium position, the restoring force acting on the combined mass $(M+m)$ is given by:
$F = -kx$

The acceleration of the system is:
$a = \dfrac{F}{M+m} = -\dfrac{kx}{M+m}$

Considering the free body diagram of the smaller block of mass $m$, the forces acting on it are the normal force $N$ upwards and its weight $mg$ downwards. Applying Newton's second law for the smaller block:
$N - mg = ma$

Substituting the expression for acceleration $a$:
$N = m(g + a)$
N = m \left(g -\dfrac{kx}{M+m}\right)
$N = mg -$\dfrac{mkx}{M+m}

To find when the normal force is smallest in magnitude, the term subtracted from $mg$ must be as large as possible. This occurs when the upward displacement $x$ is maximum.

The maximum positive displacement $x$ corresponds to the highest point of the oscillation. Therefore, the normal force is smallest at the highest point.