I assume that $(1,3)$ and $(3,1)$ are regarded as the same partition. The answer is $n=2,4$. If $n$ is odd, there is at least one odd partition while no even partition. Now consider even $n$. Given any even partition $(x_1,x_2,\ldots,x_k)$, we can construct an odd partition $(x_1-1,x_2-1,\ldots,x_k-1,1,1,\ldots,1)$. It's possible to recover $(x_1,x_2,\ldots,x_k)$ from $(x_1-1,x_2-1,\ldots,x_k-1,1,1,\ldots,1)$: first determine the number of elements $2k$, then delete $k$ copies of $1$, add $1$ to the remaining $k$ numbers. Hence this mapping is injective. But if $n \ge 6$, the odd partition $(n-3,3)$ is not mapped, so the number of odd partitions is larger than the number of even partitions.