The displacement can be positive while $\vec v$ and $\vec a$ are negative.
Imagine a car going from $x=10$ to a streetlight at $x=0$.
At every $t$ the displacement is positive since the car is always on the $x>0$ side. However surely $\vec v$ is negative since the displacement is from $x=10$ towards $x=0$, i.e. while $\vec r$ is positive, $\Delta \vec r$ is negative.
Moreover, in the case of 1d motion, it is entirely possible to have constant negative acceleration yet positive or negative velocities. This is best captured by the equation
$$
v(t)=v_0-a t\, ,\qquad a>0
$$
so that, if - say - the initial velocity $v_0= 10m/s$ while $a=5m/s^2$, we find that between $0\le t\le 2$, the velocity is positive while for $t>2$ the velocity is negative. Of course, because $v(t)$ changes sign during the motion, there will be time intervals where $\Delta \vec r>0$ and times when $\Delta \vec r<0$ since the equation for position as a function of time is (in 1d)
$$
x(t)=x_0+v_0t-\frac{1}{2}at^2= x_0+10 t -\frac{5}{2}t^2\, ,
$$
which will be positive (for $x_0=0$) for small $t$ but negative for larger $t$, i.e. the object will pass through $x=0$ at $t=0$ and $t=4$.
Actually one should not think that both $\Delta \vec v$ and $\Delta \vec a$ need to be in the same direction or even co-linear. Think of circular motion: then clearly the average velocity is (for sufficiently short times) nearly tangent to the circle but the average acceleration will be directly basically towards the centre of the circle.
A simple example is provided assuming the Moon is in circular motion about the Earth. Clearly the force $\vec F_{ME}$ will be co-linear with the acceleration but obviously here this acceleration is not in the same direction as the velocity.
