How to check does polygon with given sides' length exist?

I have polygon with $n$ angles. Then I have got $n$ values, which mean this polygon's sides' length. I have to check does this polygon exist (means - could be drawn with given sides' length). Is there any overall formula to check that? (like e.g. $a+b\ge c$, $a+c\ge b$, $c+b\ge a$ for triangle)

The only rigid polygons are triangles. If $n>3$, then there are many polygons with the same sequence of sides. This can be proved by cutting the polygon along a diagonal and using induction. The only restriction is the triangle inequality: each side is less than the sum of the other sides. Indeed, given a sequence of sides satisfying the triangle inequality you can find a triangle whose sides are two consecutives sides and the third is the remaining sides straightened flat. If you insist on angles less than $\pi$, just perturb this a bit.

No, there's no overall formula. There are some weak conditions (the sum of any $n-1$ sides' lengths must be greater than the length of the remaining side, for instance), but this is merely necessary for the existence of any polygon with those side-lengths, not one that has your desired angles. Is it sufficient? I'm not certain offhand.

Re-reading, perhaps when you said that you "have "n" angles" you meant "I'm looking for an $n$-angle polygon".

In that case, the inequalities I cited above are necessary, but are they sufficient? I suspect that they are, although they'd only guarantee a polygon with those side-lengths...not a non-self-intersecting polygon. For that latter condition, you'd have to do some additional work.

This is more a long comment to ihf's answer than a separate answer. I want to stress here that the statement made by ihf is wrong. To see why, take the side lengths $$a_1,\dots,a_n$$ all equal, with $$n \geq 5$$. Clearly $$a_1,\dots,a_n$$ satisfy the triangle inequality, but you cannot choose two consecutive side lengths $$a_i, a_{i+1}$$ (where $$a_{n+1}=a_1$$) so that there exists a triangle having as side lengths $$a_i, a_{i+1}$$ and the sum of all the remaining side lengths.

However the following is generally true. Given a sequence of positive side lengths $$a_1,\dots,a_n$$ satisfying the triangle inequality, that is $$2a_i \leq \sum_{j=1}^{n} a_j$$ for $$i=1,\dots,n$$, let us set $$a_m=a_k$$ for every integer $$m \in \mathbb{Z}$$ such that $$k \equiv m$$ (mod n). Then there exist $$m_0, m_1, m_2$$, with $$m_0 such that there is a triangle whose sides have lengths $$a_{m_0}+\dots+a_{m_1 -1}$$,$$a_{m_1}+\dots+a_{m_2 -1}$$,$$a_{m_2}+\dots+a_{m_0+n}$$.

We can easily show this by induction. For $$n=3$$ the thesis is trivial. So consider $$n \geq 4$$. We state that there are $$a_i, a_{i+1}$$ such that $$2(a_i + a_{i+1}) \leq \sum_{j=1}^{n} a_j$$. Indeed, if this were not the case, then by summing up all the inequalities $$2(a_i + a_{i+1}) > \sum_{j=1}^{n} a_j$$ for $$i=1,\dots,n$$, we would get $$4 \sum_{j=1}^{n} a_j > n \sum_{j=1}^{n} a_j$$, an absurd. Then, if $$2(a_i + a_{i+1}) \leq \sum_{j=1}^{n} a_j$$, by setting $$b_1=a_i + a_{i+1}$$, $$b_2=a_{i+2},\dots, b_{n-1}=a_{i+n-1}$$, we get a sequence of $$n-1$$ positive numbers satisfying the triangle inequality, and so the inductive hypothesis applies to give the desired triangle.

Obviously this does not give a complete proof that, given a sequence $$a_1,\dots,a_n$$ of positive numbers satisfying the triangle inequality, that is $$2a_i \leq \sum_{j=1}^{n} a_j$$ for $$i=1,\dots,n$$, there is a convex polygon having as consecutive side lengths $$a_1,\dots,a_n$$, since we have also to slightly deform the triangle found above in order to assure that each interior angle is less than $$\pi$$. This is evidently possible. Actually, Boni Bogoșel has proved muc more in his page Cyclic polygons largest area: given a sequence $$a_1,\dots,a_n$$ of positive numbers satisfying the triangle inequality, that is $$2a_i \leq \sum_{j=1}^{n} a_j$$ for $$i=1,\dots,n$$, there is a $$\mathbf{cyclic}$$ $$\mathbf{polygon}$$ having $$a_1,\dots,a_n$$ as consecutive side lengths.