Ann's claim is correct for all even $n$ greater than or equal to 18.

Denote by $A_n, B_n, C_n, D_n, E_n$ and $F_n$ the number of ways for the parcel to end at a person at a distance (in either the clockwise or anticlockwise direction) of 0, 1, 2, 3, 4, and 5 from Ann, respectively, where Ann is at a distance of 0 from herself and Jim is at a distance of 5 from Ann. Then, we have $A_0 = 1, B_0 = C_0 = D_0 = E_0 = F_0 = 0$ and for every $n \ge 0$ we get
$$A_{n+1} = 2B_n + 2C_n \quad (1)$$
$$B_{n+1} = A_n + B_n + C_n + D_n \quad (2)$$
$$C_{n+1} = A_n + B_n + D_n + E_n \quad (3)$$
$$D_{n+1} = B_n + C_n + E_n + F_n \quad (4)$$
$$E_{n+1} = C_n + D_n + E_n + F_n \quad (5)$$
$$F_{n+1} = 2D_n + 2E_n \quad (6)$$
Initial computations yield $A_1 = 0, B_1 = C_1 = 1, D_1 = E_1 = F_1 = 0$ and $A_2 = 4, B_2 = 2, C_2 = 1, D_2 = 2, E_2 = 1, F_2 = 0$.

Next assume that $n \ge 2$. Subtract (6) from (1) and use (2) to (5) to get
$$\begin{aligned} A_{n+1} - F_{n+1} &= 2(B_n + C_n) - 2(D_n + E_n) \\ &= 2(A_{n-1} + B_{n-1} + C_{n-1} + D_{n-1}) + 2(A_{n-1} + B_{n-1} + D_{n-1} + E_{n-1}) \\ &\quad - 2(B_{n-1} + C_{n-1} + E_{n-1} + F_{n-1}) - 2(C_{n-1} + D_{n-1} + E_{n-1} + F_{n-1}). \end{aligned}$$
Cancelling terms, we obtain
$$\begin{aligned} A_{n+1} - F_{n+1} &= 4(A_{n-1} - F_{n-1}) + 2(D_{n-1} - E_{n-1}) + 2(B_{n-1} - C_{n-1}) \\ &= 4(A_{n-1} - F_{n-1}) + 2(B_{n-2} - D_{n-2}) + 2(C_{n-2} - E_{n-2}) \\ &= 4(A_{n-1} - F_{n-1}) + 2(B_{n-2} + C_{n-2}) - 2(D_{n-2} + E_{n-2}) \\ &= 4(A_{n-1} - F_{n-1}) + A_{n-1} - F_{n-1} \\ &= 5(A_{n-1} - F_{n-1}). \end{aligned}$$
where we have again used equations (2) to (5). Thus, since $A_1 - F_1 = 0$ and $A_2 - F_2 = 4$ and $A_{n+1} - F_{n+1} = 5(A_{n-1} - F_{n-1})$ for every $n \ge 2$, we have
$$A_n - F_n = \begin{cases} 0 & \text{if } n \text{ is odd} \\ 4 \cdot 5^{n/2-1} & \text{if } n \text{ is even.} \end{cases}$$
Thus, for Ann's claim to be correct we need $n$ even and $4 \cdot 5^{n/2-1} \ge 10^6$, i.e., $5^{n/2-7} \ge 16$. This yields $n/2 - 7 \ge 2$ and thus $n \ge 18$.