n + S(n) + S(S(n)) = 60

Idea. A congruence cuts the candidates by a factor of $3$; a size estimate makes the list finite and short.

Congruence. $S(n)\equiv n$ and $S(S(n))\equiv n\pmod9$, so the left side is $\equiv3n\pmod9$. Since $60\equiv6\pmod9$, $3n\equiv6\pmod9$, i.e. $n\equiv2\pmod3$.

Bounds. Clearly $n<60$, so $S(n)\le S(49)=13$ or $S(59)=14$: at most $14$. Then $S(S(n))\le9$ (the maximum of $S$ over $1,\dots,14$ is $9$). Hence $n\ge60-14-9=37$.

Candidates. $n\in[37,59]$ with $n\equiv2\pmod3$: $38,41,44,47,50,53,56,59$.
- $38$: $38+11+2=51$
- $41$: $41+5+5=51$
- $44$: $44+8+8=60$ ✓
- $47$: $47+11+2=60$ ✓
- $50$: $50+5+5=60$ ✓
- $53$: $53+8+8=69$
- $56$: $56+11+2=69$
- $59$: $59+14+5=78$

Answer. $n\in\{44,47,50\}$.

Lesson. For digit-sum equations, combine the congruence $S(n)\equiv n\pmod9$ with a crude bound on $S(n)$, then check the remaining few cases.