This problem has a bijection with the one described in # 4355.

Every path in that example has a one-to-one mapping with $\{x_1$, $x_2$, $\cdots$, $x_{2n}\}$ when a value of $1$ is equivalent to a right-step and a value of $-1$ is equivalent to an up-step. Therefore, the answer is $$\boxed{\frac{1}{n+1}\binom{2n}{n}}$$

Alternatively, this problem can be solved using recursion too. Let the desired count be $a_n$.

By the definition, we must have $x_1=1$. Let $m \le n$ be the smallest integer such that $$x_1+x_2+\cdots + x_{2m}=0$$

Then $x_{2m} = -1$. This means that $x_2+x_3+\cdots +x_{2m-1}=0$ and, for any positive integer $i \le 2m-2$, it holds $x_2+\cdots + x_{i}\ge 0$. This means that the set containing $2(m-1)$ ordered numbers $\{x_2$, $x_3$, $\cdots$, $x_{2m-1}\}$ satisfies the original requirements. Therefore, there are $a_{m-1}$ such collections.

Similarly, the remaining set $\{x_{2m+1}$, $x_{2m+2}$, $\cdots$, $x_{2n}\}$ also satisfies the given requirements which means there are $a_{n-m}$ such sets.

It follows that $a_n$ satisfies $$a_n=\sum_{m=1}^{n}a_{m-1}a_{n-m}$$

This recursion can be solved using generating function (see # 4511).