Answer: does not exist. Solution: Suppose that such a function exists. We denote by $ P (m, n) $ the equality $ f (mf (n)) = f (m) f (m + n) + n $ and let $ f (1) = c.$ $ P (1,1 ): $$ f (c) = cf (2) + 1 ,$ it follows easily that $ P (1, n) : \quad f (cn) = f (n) f (n): \quad f (f (n)) = cf (n + 1) + n, \quad (1) $ $$ P(n,1): \quad f(cn)=f(n)f(n+1)+1. \quad (2)$$It follows from $(1)$ that $$ f (f (n)) \equiv n \pmod c, \quad \mbox {for any} \ n \in \mathbb {N}, \quad (3) $$If $ n \vdots c$, then using $(3)$ we obtain that $$ P (f (n), 1): \quad f (cf (n)) = f (f (n)) \cdot f (f (n) +1) +1 \equiv 1 \pmod c. $$Then $$ P (c, n): \quad f (cf (n)) = f (c) f (n + c) + n \Rightarrow f (n + c) \equiv 1 \pmod c. $$This means that $ f (n) \equiv 1 \pmod c $ for any $ n \vdots c. $ $$ P (c, n): \quad f (cf (n)) = f (c) \cdot f (n + c) + n \Rightarrow f (n + c) \equiv 1-n \pmod c, \ \mbox {for any } \ n \in \mathbb {N} \quad (4) $$Using $(4)$ for $ n = c + 2 $, it follows from (2) that $$ 1 \equiv f (c + 2) f (c + 3) +1 \equiv (-1) \cdot (-2) +1 \equiv 3 \pmod c, $$which is impossible, since $ c \ge 3, $ is a contradiction.