How many solutions does $\frac{1}{x_1}+\frac{1}{x_2}+\cdots +\frac{1}{x_n}=1$ have?

It might be worth remarking that if $G$ is a finite group with $n$ conjugacy classes and representatives $g_{1},g_{2}, \ldots ,g_{n}$, then the class equation for $G$ (divided through by $|G|$) gives $\sum_{i=1}^{n} \frac{1}{|C_{G}(g_{i})|} = 1$, though the $|C_{G}(g_{i})|$ are not usually distinct. Long ago (around 1895, I think) , E. Landau gave a bound on the number of solutions of $\sum_{i=1}^{n} \frac{1}{x_{i}} = 1$ with the $x_{i}$ (not necessarily distinct) positive integers. This allows a crude bound on the size of a finite group with $n$ conjugacy classes though this is quite far from the presently known bounds (eg of L. Pyber) which use much more group-theoretic information.

Any perfect number is an answer for your question. Also, Prof. Graham showed that $a(m)>0$ for $m>77$, where $a(m)$ denotes the total number of solution for this equation such that the sums of the denominators is $m$. You can see this page for the number of such solutions:

http://oeis.org/A051907

Also the below papers studied somehow the question:

[1] "Representation of One as the Sum of Unit Fractions" by Yuya Dan

[2] "UNIT FRACTIONS THAT SUM TO 1" by Yutaka Nishiyama

N. Burshtein has published several papers on this problem, On distinct unit fractions whose sum equals 1, and a more recent paper, making the problem more challenging by adding the restriction that the $n$ integers $x_i$ must be odd. There are no solutions for even $n$. The smallest odd $n=2k+1$ for which a solution exists is $k=4$, and the number of solutions is five.

For larger $n=2k+1$ the number of solutions (with the odd denominator restriction) is not known, but a lower bound of $(\sqrt 2)^{(k+1)(k-4)}$ was derived in Egyptian fractions with restrictions.