It depends on the surface area of the person orthogonal to the wind. I can give you an idea without getting into gross math, but I'll explain the assumptions made and where gross math would change things and why. If you have a background in vector calculus, we can run this again in a more realistic scenario.
The average adult male has a surface area of around 2m^2. We'll say that you're only facing the wind with either your front or back, and ignore the surface area from the top of your head/sides/etc. 50Kg would be a very thin "average man" but lets roll with it. Wind speed is our unknown. One assumption that we'll make is that the wind is completely orthogonal to you, so there is no vertical component. The person is standing straight up and the wind is coming and hitting them head on at a 90 degree angle
like in a wind tunnel. If that isn't the case, we just need to find the horizontal and vertical components of the force of the wind separately with some easy trig functions, but we'll quickly end up in more complicated territory if we follow that.
How much it would take to push you backwards is based on a lot of things, center of gravity, what you're standing on, what type of shoes you're wearing, but we can make an approximation again to avoid gross math. If you have a mass of 50Kg, then your weight is your mass times the acceleration due to gravity, 50Kg * 9.80m/s^2. We'll round that to 500N. For the wind to push you, it must overcome the static friction of your feet on the ground and your weight. Google says the coefficient of friction of pavement is about 0.7, and we are going to once against ignore the interface between the pavement and whatever type of shoes you're wearing. The force required to move you would be f = (coefficient of friction)*(Normal Force) where normal force is your weight. so 0.7 * 500N = 350N of force is the threshold for moving you if you stood still.
To get the force of the wind, you need the density of the medium (air) , the surface area its acting on (the person), and velocity of the wind. The density of air is 1.23kg/m^3 and we estimated the front surface area of a person to be about 1m^2. With that we can solve for the velocity squared. (sanity check, Force (Newtons) has units Kgm/s^2. If we multiply (kg/m^3) * (m^2) * ( m/s)^2 we get kgm/s^2)
So to push you back, we needed 350N. F = 1.23kg/m^3 * 1m^2 * (X)^2. X ends up being ~16.9m/s or about 37.8 miles per hour.
To actually lift that person up, you'd need the force of the wind to be greater than their weight of 500N. As mentioned about horizontal/vertical components of the force, A wind coming straight at you will never launch you up and into the air no matter how strong it is. In reality wind is a vector field and its not going to hit you with uniform velocity or direction. The best approximation for understanding the concept would be again the wind tunnel scenario above, but this time its more like one of those indoor diving simulators where you get a suit and the wind blowing up from the ground keeps you in the air. That would occur when the force of the wind is >500N or at 20.2 m/s or 45 miles per hour.
That isn't to say that a 45 mile per hour wind is going to pick you up if you are walking down the street, but its the minimum possible wind speed to lift a 50kg average man. If you turn to your side against the wind, it'll take significantly faster wind speeds to push or lift you. If you are walking forward against the wind and not just standing still, it'll impede your walking but not push you back. If you have rubber soles on your shoes, that'll change the coefficient of friction. Lots of other considerations, but I'll hold off on doing those unless you specifically want to know.
Theres a zone known as tornado alley in the southern United States where they're a pretty common occurrence. Elsewhere they're relatively rare. I don't know the exact amount of damage and fatalities tornadoes cause across the states, but they aren't a usual weather pattern. I'm reading over this now as I post it
https://www.iii.org/fact-statistic/facts-statistics-tornadoes-and-thunderstorms