Upon observation, we have a linear term on the RHS of this functional equation. Let us take $f(x) = Ax + B$ (for real integer constants $A$ and $B$) and substitute this expression into our FE: $A(m + An + B ) + B = Am + B - n$; or $Am + (A^2)n + AB + B = Am + B - n$; or $A^2n + AB = -n$; which is solvable for $B = 0, A = \pm i$. Since we require $A$ to be real and integral, no such function $f(x)$ exists for this FE for $m, n \in \mathbb{Z}$.

@above, i doult your solution is correct you just proved it for a linear function you have to prove it for any function

Let $P(m,n)$ denote the proposition $f(m + f(n)) = f(m) - n$ \begin{align*} P(0,0): f(f(0))&=f(0) \\ P(m,0): f(m+f(0))&=f(m) \\ P(m,f(0)): f(m+f(0))&=f(m)-f(0) \implies \boxed{f(0)=0} \\ P(0,m): f(f(m))&=-m \\ P(m,f(n)): f(m-n)&=f(m)-f(n) \\ P(f(-m),n): f(f(-m)+f(n))&=m-n \\ \therefore f(f(f(-m)+f(n)))=f(m-n) &\implies -[f(-m)+f(n)]=f(m)-f(n) \implies \boxed{f(m)=-f(-m)} \\ P(m,f(-n)): f(m+n)&=f(m)+f(n) \text{, hence } f \text{ is linear by Cauchy} \end{align*}Thus, let $f(m)=Am+B$. Since $f(0)=0$, $B=0$. Since $f(f(m))=-m$, $A^2 m=-m$, so no such $A$ exists. Thus, no such $f$ exists.