Mongolian Mathematical Olympiad

Let's see the given condition as a substitution $P(m, n)$. Adding to both sides $k^3$ and taking $f$ we get:

$$\left. \begin{array}{l} f(k^3 + f(m^3 + f(n))) = f(k^3 + f(m)^3 + n) \\ P(k, m^3 + f(n)) \Rightarrow f(k^3 + f(m^3 + f(n))) = f(k)^3 + m^3 + f(n) \\ \qquad \Rightarrow f(k^3 + f(m)^3 + n) = f(k)^3 + m^3 + f(n) \ (\ast). \end{array} \right\}$$

Setting in $(\ast)$ $n = l^3$ we get:
$$f(k^3 + f(m)^3 + l^3) = f(k)^3 + m^3 + f(l^3),$$
in $(\ast)$ $k = l$, $n = k^3$ we get:
$$f(l^3 + f(m)^3 + k^3) = f(l)^3 + m^3 + f(k^3)$$
Therefore,
$$f(k)^3 - f(k^3) = f(l)^3 - f(l^3) = \text{const} = c,\ c \in \mathbb{Z}$$
(1)

Setting in $(\ast)$ $k = f(s)$ we get:
$$f(f(s)^3 + f(m)^3 + n) = f(f(s))^3 + m^3 + f(n)$$
in $(\ast)$ $k = l$, $n = k^3$ we get:
$$f(f(m)^3 + f(s)^3 + n) = f(f(m))^3 + s^3 + f(n)$$
Therefore,
$$f(f(s))^3 - s^3 = f(f(m))^3 - m^3 = \text{const} = t,\ t \in \mathbb{Z}$$
In other words, $f(f(m))^3 = m^3 + t$ and for $m$ which is sufficiently large there is no perfect cube of the form $m^3 + t$, so $t = 0$. I.e. $f(f(m))^3 = m^3$ and more accurately $f(f(m)) = m$. $(\star)$

From (2) $f(f(m))^3 = m^3$
Set in (1) $l = f(m)$:
$$f(f(m))^3 = f(f(m)^3) + c$$
If we put $m = 1$ then $0 < f(f(1)^3) = 1 - c \Rightarrow c < 1$.
Therefore,
$$f(f(m)^3) = m^3 - c$$
Let's prove that $c = 0$. In the case $c < 0$:
$$f(l)^3 - c = f(l^3)$$
$P(l, -c) \Rightarrow f(l)^3 - c = f(l^3 + f(-c))$
Therefore,
$$f(l^3) = f(l^3 + f(-c))$$
On the other hand, supposing that $f(a) = f(b)$:
$$P(m, a) \Rightarrow f(m^3 + f(a)) = f(m)^3 + a \\ P(m, b) \Rightarrow f(m^3 + f(b)) = f(m)^3 + b$$
Therefore, $a = b$ which means $f$ is injective.
Moreover, $f(l^3) = f(l^3 + f(-c)) \Rightarrow f(-c) = 0$ leads to contradiction. From this follows $c = 0$.
Therefore,
$$f(1^3) = f(1)^3 \Rightarrow f(1)(f(1)^2 - 1) = 0 \Rightarrow f(1) = 1$$
And
$$P(1, n) \Rightarrow f(1 + f(n)) = 1 + n.$$
Let's prove that $f(n) = n$ by induction. Case $f(1) = 1$ is trivial. Supposing that $f(n) = n$ is true, we get $f(1 + f(n)) = f(n + 1) = n + 1$ and this completes the proof. Obviously, the function $f(n) = n$ satisfies the given condition.