Uniform distribution on $N$ and real line

I know that there is no uniform distribution on natural set since there will be contradictions with the axioms if the probability theory. In particular, if we suggest that uniform distribution exists for natural set, if $P(n) > 0$ then sum of the probabilities of elementary outcomes will be equal to infinity. If $P(n) = 0$ then sum of the probabilities of elementary outcomes will be equal to $0$. Now to the real numbers. We can uniformly define probability of choosing at random some number belonging to $[0,1]$ as $P(x) = 0$. So the question is why the sum of elementary outcomes on $[0,1] = 0 + 0 + 0 ... + 0 = 1$ while on natural set it equals to $0$?

In your question, you wrote $0+0+0+\cdots+0=1$, which is not what you meant—you meant $0+0+0+\cdots=1$, which represents the reason that there is no uniform distribution on the countable set of natural numbers. However, $0+0+0+\cdots$ inherently denotes a countable sum of $0$s, not an uncountable sum of $0$s. The probability axioms (as Brian Tung commented) only require that countable sums of disjoint probabilities add to the right thing; they don't require that of uncountable sums of disjoint probabilities.

So the facts that $P\bigl( [0,1] \bigr)=1$, and $P\bigl( \{x\} \bigr) = 0$ for all $x\in[0,1]$, and $[0,1] = \bigcup_{x\in[0,1]} \{x\}$, are all true; but they don't contradict the probability axioms, because this last union is an uncountable union, not a countable union.