Denote $g(n)=f(n)-n$. Hence $g(1)=0$ and $g(a+b+ab)=g(ab)$. Now, plugging in several values for $a,b$, we obtain \[g(2015)=g(1007)=g(503)=g(251)=g(126)=g(191)=g(95)=g(47)=g(23)=g(15)=g(7)=g(3)=g(1)=0\]Hence $\boxed{f(2015)=g(2015)+2015=2015}$

Note: While you might think that you can show $f(n) \equiv n$, in fact for all $n+1$ odd prime $f(n)$ is not forced, so $2015$ actually has a special reason here. In fact, many other values probably aren't defined either! That motivated Tintarn's solution.

Dear va2010, thank you for explaining the motivation behind my solution But actually you are wrong, since here it is shown that indeed $f(n) \equiv n$ is the only solution.

I'm pretty sure literally everyone who solved this problem at the actual contest did so by bashing out the value of $f(2015)$. Here's a solution to the general problem: We first obtain $f(n) = n$ for $n = 1, 2, 3, 4$ through brute force. We will now establish $f(n) = n$ for all $n$ by strong induction. Let $S = \{k \in \mathbb{N} \mid f(k) = k\}$, and note that the condition implies $ab + a + b \in S \iff ab \in S$. In particular taking $b = 1$ implies $2a + 1 \in S$ if and only if $a \in S$. Consider the following three cases: Case 1: $k + 1$ is composite. Then $k + 1 = (a + 1)(b + 1)$ for some positive $a, b$, thus $k = ab + a + b > ab$ and $k \in S$ since $ab \in S$ by induction hypothesis. Case 2: $k + 1$ is prime, but not a Fermat Prime. Write $k = 2^{\alpha + 1}m + 2^\alpha$ where $\upsilon_2(k) = \alpha$ and $m > 0$ as $k + 1$ is not a Fermat Prime. Then $k = 2^\alpha(2m + 1) \in S$ if and only if $2^{\alpha + 1}m + 2^\alpha + 2^{\alpha} + 2m + 1 = 2^{\alpha + 1}(m + 1) + 2m + 1$ is. This number is in $S$ if and only if $2^{\alpha}(m + 1) + m$ is in $S$, and it is straightforward to check that this number is less than $k$, and thus is in $S$. Case 3: $k + 1$ is a Fermat Prime. We can assume $k \geq 16$ thanks to our base cases. Then $k = 2^{2^m}$ for some $m \geq 2$. Note now that $$2^{2^m} = 2(2^{2^m - 1}) \in S \iff 2^{2^m} + 2^{2^m - 1} + 2 \in S$$$$2(2^{2^m - 1} + 2^{2^m - 2} + 1) \in S \iff 2^{2^m} + 2^{2^m - 1} + 2 + 2^{2^m - 1} + 2^{2^m - 2} + 1 + 2 = 2^{2^m + 1} + 2^{2^m - 2} + 5 \in S$$$$2^{2^m + 1} + 2^{2^m - 2} + 5 \in S \iff 2^{2^m} + 2^{2^m - 3} + 2 \in S$$Now note that there exists some $a$ such that $2^{2^m} + 2^{2^m - 3} + 2 = 3a + 2$, and it follows by taking $b = 2$ that this number is in $S$ if and only if $2a$ is in $S$. However it clear that $$\frac{2}{3}(2^{2^m} + 2^{2^m - 3}) < 2^{2^m}$$And so the problem is dead.

Tintarn wrote: Denote $g(n)=f(n)-n$. Hence $g(0)=0$ and $g(a+b+ab)=g(ab)$. Now, plugging in several values for $a,b$, we obtain \[g(2015)=g(1007)=g(503)=g(251)=g(126)=g(191)=g(95)=g(47)=g(23)=g(15)=g(7)=g(3)=g(1)=0\]Hence $\boxed{f(2015)=g(2015)+2015=2015}$ Actually you can't do this f(0) because N={1.2.3....}

hmida99 wrote: Actually you can't do this f(0) because N={1.2.3....} maybe he implied $g(1)=0$ not $g(0)=0$.

Ammm an ugly solution: Let $f(2a+1)=a+1+f(a)$ be Note that: $f(2015)=1008+504+252+126+63+f(62)$ But: $f(95)=48+f(47)$ $f(47)=24+f(23)$ $f(5+3+15)=8+f(15)$ $f(3)=3$ $f(3+1+3)=7$ $f(7+1+7)=f(15)=15$ $f(23)=23$ $f(47)=47$ $f(95)=95$ $f(95)=33+f(62)$ $f(62)=62$ Since that $f(2015)=2015$

That is not an ugly solution. I solved it the same way

Let $g(n)=f(n)-n$. We have $g(1)=0$ and $g(a+b+ab)+a+b+ab=a+b+ab+g(ab)$, so \[g(a+b+ab)=g(ab),\]where $g\colon \mathbb{N}\to \mathbb{Z}$. By setting $b=1$, we get that $g(2a+1)=g(a)$. We have \begin{align*} g(2015) \\ =g(1007) \\ =g(503) \\ =g(251) \\ =g(125) \\ g(62) \\ =g(6+8+6\cdot 8) \\ =g(48) \\ =g(16+3+16\cdot 3) \\ =g(67) \\ =g(33) \\ =g(16) \\ =g(2+8+2\cdot 8) \\ =g(26) \\ =g(2+13+2\cdot 13) \\ =g(41) \\ =g(20) \\ =g(4+5+4\cdot 5) \\ =g(29) \\ =g(14) \\ = g(2+7+2\cdot 7) \\ =g(23) \\ =g(3+5+3\cdot 5) \\ = g(15) \\ =g(7) \\ =g(3) \\ =g(1) \\ =0 \\ \end{align*} So $f(2015)=g(2015)+2015=\boxed{2015}$.