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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 $n=9$.