A digital feedback network question
This is tricky. There are two things to consider:


Can the system be implemented as depicted?
Can the difference equation described by the system (i.e., the input/output relation) be implemented?


The answer to the first question is indeed 'no'. You can't implement the system with the structure given in the figure. The answer to the second question is 'yes'. I'll show you what I mean.

Defining two (time dependent) auxiliary variables $v[n]$ and $w[n]$ at the outputs of the two adders at the top left of the figure, respectively, the system is described by the following set of difference equations:

$$\begin{align}v[n]&amp;=x[n]+G(x[n]-y[n])\\
w[n]&amp;=v[n]+Aw[n-1]+Cw[n-2]\\
y[n]&amp;=w[n]+Bw[n-1]+w[n-2]\end{align}\tag{1}$$

As you've noticed, these equation cannot be implemented directly, because in order to compute $y[n]$ you need $v[n]$, but in order to compute $v[n]$ you need $y[n]$. However, these equations can be 'unrolled' and rewritten in such a way that they can be implemented. This can be shown by using the $\mathcal{Z}$-transform of $(1)$ to convert difference equations into algebraic equations:

$$\begin{align}V(z)&amp;=X(z)(1+G)-GY(z)\\
W(z)&amp;=V(z)+W(z)(Az^{-1}+Cz^{-2})\\
Y(z)&amp;=W(z)(1+Bz^{-1}+z^{-2})\end{align}\tag{2}$$

Now we can eliminate $V(z)$ and $W(z)$ from these three equations, and with a bit of algebra we arrive at

$$H(z)=\frac{Y(z)}{X(z)}=\frac{(1+G)(1+Bz^{-1}+z^{-2})}{1+G+(GB-A)z^{-1}+(G-C)z^{-2}}\tag{3}$$

For $G\neq -1$, Eq. $(3)$ corresponds to the following difference equation:

$$y[n]=x[n]+Bx[n-1]+x[n-2]+\frac{A-GB}{1+G}y[n-1]+\frac{C-G}{1+G}y[n-2]\tag{4}$$

which can be implemented without any problem. Note that $(4)$ implements the same difference equation as described by the depicted system, but it uses another structure such that there are no delay-free loops.

To help your intuition, throw away the block with the two delay elements including the multipliers and the two adders, and just consider the trivial system described by

$$y[n]=x[n]+G(x[n]-y[n])\tag{5}$$

There is a delay-free loop, but if $G\neq-1$ the system can be implemented without any problem by rewriting $(5)$ as

$$y[n](1+G)=x[n](1+G)\tag{6}$$

which (under the mentioned assumption on $G$) is of course equivalent to $y[n]=x[n]$. This principle also applies to much more complicated systems.