Croatian Mathematical Competitions

Observe that $n = 1$ is not a solution, so $d(n) \ge 2$.
It is impossible that $d(n) = 2$, since then $n$ would be prime and given equation would reduce to $1 + n = n + 2 + 1$, so $d(n) \ge 3$.
Let $1 = D_1 < D_2 < \cdots < D_d = n$ be the divisors of $n$. The given equation can then be written as
$$1 + \sum_{i=2}^{d(n)-1} D_i + n = n + d(n) + 1.$$
Since $D_i \ge 2$ for all $i \in \{2, 3, \dots, d(n) - 1\}$, we get $d(n) = \sum_{i=2}^{d(n)-1} D_i \ge (d(n) - 2) \cdot 2$,
hence $d(n) \le 4$, i.e. $3 \le d(n) \le 4$.
For $d(n) = 3$, number $n$ is a square of some prime $p$ and given equation reduces to
$$1 + p + p^2 = p^2 + 3 + 1,$$
hence $p = 3$ and $n = p^2 = 9$.
For $d(n) = 4$, there are two possibilities:
a) If $n$ is product of some primes $q$ and $r$ ($q < r$), then the given equation implies $1 + q + r + qr = qr + 4 + 1$, thus $q + r = 4$. There are no such primes $q$ and $r$.
b) If $n$ is a cube of some prime $s$, then the given equation means that $1+s+s^2+s^3 = s^3 + 4 + 1$, thus $s^2 + s - 4 = 0$. There is no such prime $s$.

Therefore, the only solution is $n = 9$.