A metric on the natural numbers

Yes, there are many. For example, consider the metric $d(m, n)=\vert {1\over 2^m}-{1\over 2^n}\vert+1$ for $m\not=n$, and $d(m, m)=0$. Given any $n$, the set $\{n+k: k\in\mathbb{N}\}$ is the closed ball of radius $1+{1\over 2^{n+1}}$ centered at $n+1$. Meanwhile, it's trivially complete: since every distinct pair of points are at distance at least 1, there are no non-eventually-constant Cauchy sequences.

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Note that if $M=(X, d)$ is a metric space, then $M^+=(X, d^+)$ is also a metric space, where $d^+(a, b)=d(a, b)+1$ for $a\not=b$ and $d(a, a)=0$. The only nontrivial axiom to check is the Triangle Inequality: if $a, b, c$ are distinct, then $$d^+(a, b)+d^+(b, c)=[d(a, b)+d(b, c)+1]+1\ge [d(a, c)+1]+1>d^+(a, c).$$ (If $a=b$ or $b=c$, things are even easier.) This construction can be a useful source of counterexamples.

Note that this is just a special case of the sum of two metrics: if $d_0, d_1$ are metrics on $X$, then so is $d(x, y)=d_0(x, y)+d_1(x, y)$. Here we're taking the sum of a metric which satisfies the hypotheses of your question except for completeness, and a discrete metric (which trivializes the completeness requirement while preserving the other relevant property).