If $$$1-\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{5+\frac{1}{6+\frac{1}{7}}}}}}=\frac{1}{x_1+\frac{1}{x_2+\frac{1}{x_3+\frac{1}{x_4+\frac{1}{x_5+\frac{1}{x_6+\frac{1}{x_7}}}}}}}$$$ where $$x_{1},x_{2},....,x_{7}$$ are positive integers, find $$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}$$.

We first evaluate the left‐hand expression bottom-up.

Start with the innermost term:
$$6+\frac1{7}=\frac{43}{7}$$

Move one level outward each time:

$$5+\frac1{\displaystyle 6+\frac1{7}} =5+\frac{1}{\tfrac{43}{7}} =5+\frac{7}{43} =\frac{222}{43}$$

$$4+\frac1{\displaystyle 5+\frac1{6+\frac1{7}}} =4+\frac{1}{\tfrac{222}{43}} =4+\frac{43}{222} =\frac{931}{222}$$

$$3+\frac1{\displaystyle 4+\frac1{5+\cdots}} =3+\frac{1}{\tfrac{931}{222}} =3+\frac{222}{931} =\frac{3015}{931}$$

$$2+\frac1{\displaystyle 3+\frac1{4+\cdots}} =2+\frac{1}{\tfrac{3015}{931}} =2+\frac{931}{3015} =\frac{6961}{3015}$$

Finally,

$$1-\frac1{\displaystyle 2+\frac1{3+\cdots}} =1-\frac{1}{\tfrac{6961}{3015}} =1-\frac{3015}{6961} =\frac{3946}{6961}$$

Hence

$$1-\frac{1}{2+\frac{1}{3+\cdots}}=\frac{3946}{6961} =\frac1{x_1+\frac1{x_2+\frac1{x_3+\frac1{x_4+\frac1{x_5+\frac1{x_6+\frac1{x_7}}}}}}}$$

Taking reciprocals gives the ordinary continued fraction we need:

$$x_1+\frac1{x_2+\frac1{x_3+\frac1{x_4+\frac1{x_5+\frac1{x_6+\frac1{x_7}}}}}} =\frac{6961}{3946}$$

Now expand $$\frac{6961}{3946}$$ as a simple continued fraction using the Euclidean algorithm:

$$\frac{6961}{3946}=1+\frac{3015}{3946}=1+\frac1{\tfrac{3946}{3015}} \quad\Rightarrow\; x_1=1$$
$$\frac{3946}{3015}=1+\frac{931}{3015}=1+\frac1{\tfrac{3015}{931}} \quad\Rightarrow\; x_2=1$$
$$\frac{3015}{931}=3+\frac{222}{931}=3+\frac1{\tfrac{931}{222}} \quad\Rightarrow\; x_3=3$$
$$\frac{931}{222}=4+\frac{43}{222}=4+\frac1{\tfrac{222}{43}} \quad\Rightarrow\; x_4=4$$
$$\frac{222}{43}=5+\frac{7}{43}=5+\frac1{\tfrac{43}{7}} \quad\Rightarrow\; x_5=5$$
$$\frac{43}{7}=6+\frac{1}{7}=6+\frac1{\tfrac{7}{1}} \quad\Rightarrow\; x_6=6$$
$$\frac{7}{1}=7 \quad\Rightarrow\; x_7=7$$

Thus
$$x_1=1,\;x_2=1,\;x_3=3,\;x_4=4,\;x_5=5,\;x_6=6,\;x_7=7.$$

The required sum is
$$x_1+x_2+x_3+x_4+x_5+x_6+x_7 =1+1+3+4+5+6+7=27.$$

Answer: 27