Answer: 799

Small Hint:

For $N=6^{20},$ rearrange $\frac{2xy}{x+y}=N$ into a product

Big Hint:

Count complementary factor pairs of $N^2$ in which both factors are even

Solution:

Let $N=6^{20}.$ Rearranging the harmonic-mean equation gives $$(2x-N)(2y-N)=N^2.$$ Because $x<N<y$ and $N<2x,$ the two factors are positive, and both must be even. Conversely, each factorization $AB=N^2$ with even $A<B$ gives one valid pair via $x=\frac{A+N}{2}$ and $y=\frac{B+N}{2}.$

Now $N^2=2^{40}3^{40}.$ For both complementary factors to be even, the exponent of $2$ in $A$ can be $1,\ldots,39,$ while the exponent of $3$ can be $0,\ldots,40.$ This gives $39\cdot41=1599$ divisors $A,$ including the central factor $A=N.$ Pairing complementary divisors and excluding that central case gives $\frac{1599-1}{2}=799.$