The smallest and the largest possible values of $R$ are $0$ and $\frac{n(n+1)(2n+1)}{6}$ respectively.
Let the points be $A_1, A_2, \dots, A_{2n+1}$. If we draw the vector $\overrightarrow{A_iA_j}$ whenever $i < j$, then for any $\triangle A_iA_jA_k$ with $i < j < k$, we have
$$\overrightarrow{A_iA_j} + \overrightarrow{A_iA_k} + \overrightarrow{A_jA_k} = 2\overrightarrow{A_iA_k} \neq \mathbf{0}.$$

Therefore, it is possible that $R = 0$.
For the maximum value, we call $\triangle A_i A_j A_k$ special if it does not have sum $0$. Note that the vectors of a special triangle must be $\overrightarrow{A_i A_j}$, $\overrightarrow{A_i A_k}$ and $\overrightarrow{A_j A_k}$ up to symmetry. We say that $A_i$ is the initial vertex and $A_k$ is the terminal vertex of this triangle.
For each vertex $A_i$, suppose there are $b_i$ vectors pointing away from $A_i$ and $c_i$ vectors pointing towards $A_i$. Note that $b_i + c_i = 2n$. Then there are $\binom{b_j}{2}$ pairs of vectors $\overrightarrow{A_i A_j}$ and $\overrightarrow{A_i A_k}$, and $\binom{c_j}{2}$ pairs of vectors $\overrightarrow{A_j A_i}$ and $\overrightarrow{A_k A_i}$. Therefore, there are $\binom{b_j}{2}$ special triangles with $A_i$ as initial vertex (note that the vector between $A_j$ and $A_k$ is irrelevant), and $\binom{c_j}{2}$ special triangles with $A_i$ as terminal vertex. As each special triangle is counted twice, it follows that the number of special triangles is
$$\frac{1}{2} \sum_{j=1}^{2n+1} \left[ \binom{b_j}{2} + \binom{c_j}{2} \right] \ge \frac{1}{2} \sum_{j=1}^{2n+1} \left[ \binom{n}{2} + \binom{n}{2} \right] = \frac{n(n-1)(2n+1)}{2}$$
by Jensen's inequality. Since there are $\binom{2n+1}{3}$ triangles in total, this implies
$$R \le \binom{2n+1}{3} - \frac{n(n-1)(2n+1)}{2} = \frac{n(n+1)(2n+1)}{6}.$$
This maximum value $R = \frac{n(n+1)(2n+1)}{6}$ can be attained. For example, suppose we draw the vector $\overrightarrow{A_j A_{j+m}}$ for $1 \le j \le 2n+1$ and $m = 1, 2, \dots, n$, where the indices are taken modulo $2n+1$. Then $b_j = c_j = n$ for any $j$, and so equality holds.