= Solution
Put $R=\epsilon+\delta$. Only the first row disc changes: it now has radius $R$. The other discs keep radii $r_i=1/4+\epsilon$ for $2\le i\le n-1$ and $r_n=1/4$. They remain mutually disjoint. As $R$ grows, the first disc joins an initial segment of these discs. Define $m$ as the largest index joined to it, or one if no other disc is joined. The joining condition for index $i$ is
$$
R+r_i\ge i-1.
$$
For an interior index this starts at $\delta=i-5/4-2\epsilon$, and for the final index at $\delta=n-5/4-\epsilon$. Until its joining threshold, disc $i$ contains a unique real <eigenvalue> and retains the bound $|\lambda_i-i|\le r_i$.
After the first $m$ discs have joined, the second theorem places exactly $m$ eigenvalues in their union and exactly one in each remaining disc. The cluster is bounded, for instance, by
$$
|z-1|\le\max\{R,m-1+r_m\}\quad(m\ge2),
$$
and by $|z-1|\le R$ when $m=1$. The union of the actual discs is a sharper enclosure. Once discs overlap, one cannot assign one <eigenvalue> to each individual disc merely by its centre; the theorem counts the whole separated cluster.
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