f is a function from ${\mathbb{Z}}^{+}$ to  ${\mathbb{Z}}^{+}$, where ${\mathbb{Z}}^{+}$ is the set of non-negative integers which has the following properties :  

a) $f(n+1)>f\left(n\right)$  for each $n\in {\mathbb{Z}}^{+}$ 
b) $f(n+f\left(m\right))=f\left(n\right)+m+1$ for all $m,n\in {\mathbb{Z}}^{+}$  
The function satisfies the conditions (a) & (b) and yields a unique value of $f\left(2023\right)$ is a four digit number ABCD then A+B+C+D is equal to 

1. Let  $f\left(0\right)=k$, at $m=0$, where  k is a non – negative integer 
Then the condition b) gives 
$f(n+k)=f\left(n\right)+1$-----(1) 
 Putting  $n=0$ in (1), we get  $f\left(k\right)=f\left(0\right)+1$
    $\Rightarrow f\left(k\right)=k+1------\left(2\right)$
Substituting $n-1$ for n and k form into condition (b) we get
$$\begin{gathered} f(n-1+f\left(k\right))=f(n-1)+k+1 \\ \Rightarrow f(n-1+k+1)=f(n-1)+k+1 from \left(2\right) \\ \Rightarrow f(n+k)=f(n-1)+k+1 -----\left(3\right) \end{gathered}$$

From (1) & (3)  $f\left(n\right)+1=f(n-1)+k+1$
 $\Rightarrow f\left(n\right)=f(n-1)+k$
$\forall n\in {\mathbb{Z}}^{+}$  using this repeatedly gives 
$f\left(n\right)=f(n-1)+k$

$$\begin{gathered} =[f(n-2)+k]=(n-2)+2k \\ =f(n-3)+3k= \dots \dots \dots = f\left(0\right)+nk \\ If f\left(n\right)=(n+1)k then f\left(0\right)=k \end{gathered}$$

From (2),  $f\left(k\right)=k+1,$ so that we have 

$(k+1)=k(k+1) \Rightarrow k=\pm 1$

$\because f\left(0\right)=k,$  is a non- negative integer,  $k=1$
Hence  $f\left(2023\right)=(2023+1)\times 1$
$=2024$  ABCD and A+B+C+D=8
Note that condition $\left(a\right)$ is satisfied by this value of k.