Let z be the complex number with $\operatorname{Im}(\mathrm{z})=10$ and satisfing $\frac{2 \mathrm{z}-\mathrm{n}}{2 \mathrm{z}+\mathrm{n}}=2 i-1$, where $i=\sqrt{-1}$, for some natural number ' n ' then

The complex number is given by $z = x + 10i$, since $\operatorname{Im}(z) = 10$. Substituting $z$ into the equation:

$\frac{2(x + 10i) - n}{2(x + 10i) + n} = 2i - 1$

Simplify the numerator and denominator:

$\frac{(2x - n) + 20i}{(2x + n) + 20i} = -1 + 2i$

Multiply both sides by the denominator:

$(2x - n) + 20i = (-1 + 2i)((2x + n) + 20i)$

Expand the right-hand side:

$(2x - n) + 20i = -1(2x + n) - 20i + (4x + 2n)i + 40i^2$

Since $i^2 = -1$:

$(2x - n) + 20i = -2x - n - 20i + (4x + 2n)i - 40$

Group real and imaginary parts:

Real: $2x - n = -2x - n - 40$
Imaginary: $20 = -20 + 4x + 2n$

From the real equation:

$2x = -2x - 40 \Rightarrow 4x = -40 \Rightarrow x = -10$

Substitute into the imaginary equation:

$40 = 4(-10) + 2n \Rightarrow 40 = -40 + 2n \Rightarrow 2n = 80 \Rightarrow n = 40$

The values are $n = 40$ and $\operatorname{Re}(z) = -10$, corresponding to option D.