Say the weights $(p_k)$ yield the asymptotic distribution $(q_k)$. The one-step Markov dynamics of the process means that, for every $k$ between $1$ and $N$, $ q_k=\sum_{i=1}^Nq_i\frac{p_k}{P_{i+1}}[i+1\ge k],\quad\text{where}\ P_k=\sum_{i=1}^kp_i, $ with the convention that $P_{N+1}=1$.
If a collection of weights $(p_k)$ yields the uniform distribution $q_k=1/N$ for every $k$ between $1$ and $N$, the asymptotic mean will be $\frac12(N+1)$, as required. But this happens if and only if, for every $k$, $ \frac1{p_k}=\sum_{i=1}^N\frac1{P_{i+1}}[i+1\ge k]. $ In particular, $ \frac1{p_1}=\sum_{i=1}^N\frac1{P_{i+1}}=\frac1{p_2}, $ hence $p_1=p_2=x$, say, and $P_1=x$, $P_2=2x$. Likewise, $ \frac1{p_3}=\sum_{i=2}^N\frac1{P_{i+1}}=\frac1{p_2}-\frac1{P_2}=\frac1{x}-\frac1{2x}, $ hence $p_3=2x$ and $P_3=4x$. From here, using the relations $ \frac1{p_k}=\frac1{p_{k-1}}-\frac1{P_{k-1}},\quad P_k=P_{k-1}+p_k, $ one proves recursively that $p_k=2^{k-2}x$ and $P_k=2^{k-1}x$ for every $k$ between $2$ and $N$. Since $P_N=1$, this yields $x=2^{1-N}$.
Finally, a suitable collection of weights $(p_k)$ is $ p_1=2^{1-N},\quad p_k=2^{k-1-N},\quad 2\le k\le N. $