jmc

Let $x$ be the given number, so that $x=2+\frac{4}{1+\frac{4}{\left(2+\frac{4}{1+\cdots }\right)}}$. The term in the parentheses is exactly the definition of $x$, so it follows that $$x=2+\frac{4}{1+\frac{4}{x}}=2+\frac{4x}{x+4}.$$ Multiplying by $(x+4)$ on both sides and simplifying gives $x(x+4)=2(x+4)+4x⟹x^2+4x=2x+8+4x.$ Thus, we have the quadratic equation $$x^2-2x-8=(x-4)(x+2)=0,$$ and it follows that $x=-2,4$. Since the given number is positive, the answer is $\boxed{4}$.