Do there exist several positive real numbers such that their sum is $1$ and sum of their squares is less than $0.01$

Just take positive reals $x_1,x_2,\dots ,x_n$ so that their sum is $1$ and all the numbers are smaller than $\frac{1}{100}$.

Then we have: $x_1^2+x_2^2+\dots x_n^2<\frac{1}{100}(x_1+x_2+\dots + x_n)=\frac{1}{100}\cdot1=0.01$

Yes. Take $x_n=\frac{\epsilon}{n}$, where $\epsilon=\left(\sum_{n=1}^N\frac{1}{n}\right)^{-1}$. We have $$\sum_{n=1}^Nx_n=\epsilon\sum_{n=1}^N\frac{1}{n}=1$$ by definition. On the other hand, $$\sum_{n=1}^Nx_n^2=\frac{\sum_{n=1}^N\frac{1}{n^2}}{\left(\sum_{n=1}^N\frac{1}{n}\right)^2}.$$ By taking $N$ large enough, this can be arbitrarily small since the numerator converges, while the denominator diverges to $\infty$.

Suppose all the numbers will be equal. Then we have:

$$n \times i^2 < 0.01$$ $$n \times i = 1$$

where $i$ is the number we're looking for and $n$ is the quantity. Substitute and get:

$$i < 0.01$$