= Solution
Write $Q=\sum_{m=1}^N1/m$ and propose $\pi_N=1/Q$, $\pi_j=1/((j+1)Q)$ for $j<N$. These probabilities sum to one because $1+\sum_{j=1}^{N-1}1/(j+1)=Q$.
For $j<N$, the incoming stationary <probability> is
$$
\sum_{k=j+1}^N\frac{\pi_k}{k}
=\frac1Q\left\{\sum_{k=j+1}^{N-1}\frac1{k(k+1)}+\frac1N\right\}
=\frac1{(j+1)Q}=\pi_j,
$$
using the telescoping identity $1/(k(k+1))=1/k-1/(k+1)$. Similarly $\sum_{k=1}^N\pi_k/k=1/Q=\pi_N$. Thus $\pi$ is stationary.
The finite chain is irreducible: every state can reach $N$, and from $N$ every state is reachable in one step. Its self-loop at $N$ makes it aperiodic. The finite-state convergence theorem therefore gives the claimed limit from any initial state:
$$
\boxed{\lim_{n\to\infty}\Pr(X_n=j)=\begin{cases}1/((j+1)Q),&1\leq j<N,\\1/Q,&j=N.\end{cases}}
$$
For $N=1$ this is the single absorbing refill state and the same formula gives <probability> one.
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