The comment by mlk is right on the mark. The idea of this solution is basically what he wrote, but there a lot of messy details coming from the question of which direction the arcs are curving. Let the points be $P_1$, $P_2$, ... $P_n$ in that order, and assume that the curve is a simple closed curve. Let $\alpha_i$ be the angle $\angle(P_{i-1} P_i P_{i+1})$, measured inside the simple closed polygon. I will show that:
For $n=2m$ even, the curve closes if and only if $\alpha_1+\alpha_3+\cdots+\alpha_{2m-1} = \alpha_2 + \alpha_4 + \cdots + \alpha_{2m}$. In this case, we can start at any angle. (Since the sum of the angles of a $2m$-gon is $(2m-2) \pi$, the common value of the sum of $(m-1) \pi$.)
If $n$ is odd, there is exactly one angle we should start at.
Let $O_i$ be the center of the arc from $P_i$ to $P_{i+1}$. I found the following sign convention useful: If $O_i$ is inside the curve, I set $\theta_i = \angle(O_i P_i P_{i+1}) = \angle(O_i P_{i+1} P_i)$; these angles are equal because they are the base angles of an isosceles triangle. If $O_i$ is outside the curve, I put $\theta_i = \pi-\angle(O_i P_i P_{i+1}) = \pi-\angle(O_i P_{i+1} P_i)$.
The condition that the arcs $P_{i-1} P_i$ and $P_i P_{i+1}$ are tangent says that $O_{i-1}$, $O_i$ and $P_i$ are collinear with $O_{i-1}$ and $O_i$ on the same (respectively opposite) side of $P_i$ if $O_i$ and $O_{i+1}$ are on the same (respectively opposite) side of the curve.
I claim that the following relation holds:
$$\alpha_i = \theta_{i-1} + \theta_i. \qquad (\ast)$$
We need to check all four cases for whether $P_{i-1}$ and $P_i$ are inside or outside of the curve: For example, if both are inside, then this holds because
$$\angle(P_{i-1} P_i P_{i+1}) = \angle(P_{i-1} O_{i-1} P_i)+ \angle(P_{i} O_{i} P_{i+1}) = \angle(P_{i-1} O_{i-1} P_i)+ \angle(P_{i} O_{i+1} P_{i+1}).$$
In the other cases, we need to similarly check that all the $\pi$'s and $-$'s work out.
We can use $(\ast)$ to compute all of the $\theta_i$ in terms of $\theta_1$: We have $\theta_2 = \alpha_2-\theta_1$, $\theta_3 = \alpha_3-\alpha_2+\theta_1$, etcetera.
If $n$ is even, we see that the curve closes if and only if $\alpha_1+\alpha_3+\cdots+\alpha_{2m-1} = \alpha_2 + \alpha_4 + \cdots + \alpha_{2m}$. If $n$ is odd, we see that the key equation is $\theta_1 = \alpha_1 - \alpha_2+\alpha_3-\alpha_4 +\cdots +\alpha_n - \theta_1$. This will hold exactly for one value of $\theta_1$ modulo $\pi$. $\square$
I tried to work out the case of a self intersecting curve, but it got too messy. There is a method called "directed angles" which often cleans up situations like this -- see Section 1.7 of Geometry Unbound -- but I didn't find it helpful.