I dont know if this can be taken as a totally correct solution but : by observation I can see that a function satisfying the given condition is f(k) = k+95 (but i have to prove that it is unique...how?)Then the value of the sum is 1995

We have $ f(f(m) + f(n)) = n + f(f(m) + 95) = m + n + f(2.95)\forall m,n\in\mathbb{N}$ and if $ f(m) = f(n)$ then $ m = n$ therefore $ f(n + 1) + f(n - 1) = 2f(n)\forall n > 1$. Done!

Today i've known a nice proof for this problem:, i will replace $95$ by $K$: - First, $1 + f(m + f(n)) = n + 1 + f(m + K) \implies f(1 + f(m + f(n))) = f(n + 1 + f(m + K)) \implies m + f(n) + f(K + 1) = m + K + f(n + K + 1) \implies f(n + K + 1) = f(n) + f(K + 1) - K$ (*) - Second, $P(m, n) is f(m + f(n)) = n + f(m + K)$: i) $P(m, K + f(n)): f(m + n + f(2K)) = f(m + f(K + f(n))) = f(n) + K + f(m + K)$ ii) Change $m, n$ in i), $f(m + n + f(2K)) = f(m) + f(n + K) + K$ Therefore, $f(m + K) + f(n) + K = f(m) + f(n + K) + K \implies f(m + K) - f(m) = f(n + K) - f(n) \implies f(n + K) - f(n) = f(K + 1) - f(1)$ (**) $$P(m, n + 1) in (**): f(n + K + 1) - f(n + 1) = f(K + 1) - f(1) \implies f(n) + f(K + 1) - K - f(n + 1) = f(K + 1) - f(1) \implies f(n + 1) - f(n) = f(1) - K$$ $$\implies f(n) - f(1) = (n - 1)(f(1) - K) \implies f(n) = n(f(1) - K) + K $$ Use it in the first, we have: (Set $a = f(1)$) $(m + f(n))(a - K) + K = n + (m + K)(a - K) + K$ $\implies f(n)(a - K) = n + K(a - K)$ $\implies (n(a - K) + K)(a - K) = n + Ka - K^2$ $\implies n(a - K)^2 = n$ So $a = K + 1$ or $a = K - 1$ Case $a = K + 1$ we get $f(n) = n + K$ Case $a = K - 1$ we get $f(n) = K - n$, let $n := 2K$ then f(2K) < 0, contradition. So there exist a unique function $f$. Done. P.s: Sorry for bump up this topic. I want to share nice idea .