Can smoke stay still in the air? Can a small amount of smoke be dense enough to stay in the air keeping its shape for a minute or so?
Or does it always dissipate quickly?
If not smoke, can anything else stay in the air for a minute while keeping its shape?
 A: 
Can a small amount of smoke be dense enough to stay in the air keeping its shape for a minute or so?
  Or does it always dissipate quickly?

I read this and think to myself "optimization problem".  Firstly, you should know the following, which is the law of diffusion:
$$\frac{\partial \phi}{\partial t} = D\,\frac{\partial^2 \phi}{\partial x^2}$$
For clarity, $\phi$ is a function that represents the distribution of the concentration of the gas.  It is a scalar function of 3 variables, which is to say that I could write it as $\phi(\vec{r})$, where $\vec{r}$ represents typical $x,y,z$ coordinates.  If have an image in your mind of a cloud that is the shape of a snowman, that can be represented by that function, so can smoke rings or whatever you desire.
The clarity of the shape degrades over time, exactly per the above diffusion equation.  Picture blurring an image in Photoshop.  That is very similar to the process that happens.


*

*Q: Can you reduce the rate at which this blurring happens?

*A: Yes you can


The rate at which $\phi$ (your snowman) degrades in sharpness comes from the magnitude of $|d\phi/dt|$.  This magnitude is proportional to the diffusion coefficient $D$ as well as that other derivative with respect to $dx^2$, but that term is representative of the sharpness itself, so we don't want to reduce that, we would rather reduce $D$.  In order to reduce $D$, we need to first talk about mean free path and velocity of the gas molecules.  I'll use this source and refer to the mean free path $\lambda$ (units of length) and average speed $\bar{c}$.  In general D is proportional to those two.
$$D \propto \lambda \bar{c}$$
For a gas cloud the parameter $\lambda$ has mostly to do with the density of the gas, as well as some other things.  Again, we would like to minimize $D$, but $\lambda$ might not have much design freedom.  On the other hand, $\bar{c}$ could have great design freedom.  This is also dependent on the temperature of the gas, but more specifically, the temperature is a measure of the kinetic energy of the molecules.  I'll say fairly generally:
$$\frac{1}{2} m \bar{c}^2 = \frac{3}{2} k T$$
Never mind very much what $k$ is (it's just a physical constant), what matters is that this equation has temperature $T$ and $m$.  I'm taking your question to be most likely concerned with normal air.  That means that it is unlikely that we would have $T$ as a design variable.  However, since you are not specifying the gas you are working with, it's possible we could choose that, and the selection of that gas determines $m$ which is a factor in determining $\bar{c}$ which is a factor in determining $D$, which determines the persistence of your cloud image.
Bottom line: Heavier gases will diffuse more slowly, meaning the image will persist longer.
An example of a high molecular weight gas is common refrigerant gases, like R-134a.  If you released that into the air it will diffuse rather slowly compared to other examples.  NOTE: don't do this, it would be dangerous and probably illegal.
A: Quick answer: I believe due to Brownian Motion, a small cloud of smoke particles would dissipate in fairly short order. Indeed, smoke particles are used in experiments to confirm the particulate nature of Air.
However, if you were to wrap that smoke in a soap bubble, the surface tension could hold for longer than a minute…
A: Would, say, clouds or fog count? In this case, the temperature differences between pockets of the air leads to accumulation of water vapor. Off the top of my head, it might be possible to imagine a Van der Waals stabilized gas, or a heterogenous electrostatically stabilized 'smoke'.
