Why is a particle non-relativistic when its kinetic energy is small compared to its rest energy?

I would like to add something to the already great answers posted.

Obviously, non-relativistic is a qualitative term, you can translate it to "relativistic effects are so small that they're negligible in this problem".

In the particular case you're talking about, and as was pointed out by Roger JBarlow and John Rennie, you can calculate the Lorentz factor to be $\gamma=1.01$. This means you are going to have measurement errors on the order of $10^{-2}$. In some fields this may be acceptable (it would be beyond amazing in fluid mechanics) , but I recall a great professor I had on relativity (he works in numerical relativity, and is one of the leading figures on the field, at least in my country) who said "If the errors are on the order of $10^{-4}$, the results are basically useless". This is further illustrated by the fact that accurate GPS measurements rely on accurate calculation of relativistic effects which are (if I recall correctly) on the order of $10^{-12}$, and would otherwise give errors of kilometers.

The bottom line is that the question "is this particle non-relativistic?" Is basically the same as asking "is $\gamma$ close enough to 1 so that I can just assume it's 1?". This will change depending on the problem under consideration.

When we say a particle is non-relativistic we mean the Lorentz factor $\gamma$ is close to one, where $\gamma$ is given by:

$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$

So saying $\gamma$ is close to one means that the velocity $v$ must be much less than $c$.

With a bit of algebra we can show that the kinetic energy of a particle is given by:

$$ T =(\gamma - 1)mc^2 $$

And the rest mass energy is the usual $mc^2$, so if we take the ratio of the kinetic energy to the rest mass energy we get:

$$ \frac{T}{E} = \frac{(\gamma - 1)mc^2}{mc^2} = \gamma - 1 $$

And if this ratio is small that means $\gamma \approx 1$, which was our original criterion for non-relativistic behviour.

'Non-relativistic' means $v\ll c$.

That is effectively the same as $\gamma \approx 1$ as $\gamma={1 \over \sqrt{1-v^2/c^2}}$.

But also $\gamma={E_{tot}\over E_{rest}} \equiv 1+{E_{kin} \over E_{rest}}$

So if the kinetic energy is small compared to the rest mass, $\gamma$ is only slightly bigger than 1, and $v/c$ is small. And one is justified in ignoring relatistic effects.