The second condition implies that for all $a, b \in \mathbb{N}$ the triangle inequalities hold:
$$\begin{aligned} f(a) + f(b) &> f(a+b-1), \\ f(a) + f(a+b-1) &> f(b), \\ f(b) + f(a+b-1) &> f(a). \end{aligned}$$
First, consider $a = b = 2$. It then follows from the first condition that $f(4) = f(2)^2$, and the second condition implies that $2f(2) > f(3)$.

Now, let $a = 3$, $b = 2$. The second condition then implies that $f(2) + f(3) > f(4)$. So,
$$\begin{aligned} f(2)^2 &= f(4) < f(2) + f(3) < f(2) + 2f(2) = 3f(2) \\ &\implies f(2)(3 - f(2)) > 0. \end{aligned}$$
Since $f(2)$ is a positive integer there are only two possibilities: $f(2) = 1$ and $f(2) = 2$. We consider the two cases separately:

* If $f(2) = 1$, then let $a = 2$, $b = 1$. The second condition implies that $2f(2) > f(1)$, so $f(1) = 1$. We can now use induction to show that $f(n) = 1$ for all $n$. This is true for the base case. Now, assume that $f(n) = 1$ for some $n \ge 2$. Using $a = n$, $b = 2$ in the second condition we get
$$f(n+1) < f(n) + f(2) = 2 \implies f(n+1) = 1.$$
This gives us the solution $f(n) = 1$ for all $n \in \mathbb{N}$.

* If $f(2) = 2$, then $f(4) = f(2)^2 = 4$. What is more, for all $k \in \mathbb{N}$ the induction on the first condition implies that
$$f(2^k) = f(2)f(2^{k-1}) = \dots = f(2)^k = 2^k.$$
We know from before that $f(4) - f(2) < f(3) < 2f(2)$, which implies that $f(3) = 3$.

We now use induction to show that $f(n) = n$ for all $n \ge 2$. The base case is obvious. Now, assume that the induction hypothesis holds for $2, 3, \dots, n-1$. Using $a = n-1$, $b = 2$ we get
$$f(n) < f(n-1) + f(2) = n+1 \implies f(n) \le n.$$
Let $2^r$ be the greatest power of $2$, such that $2^r \le n$. If $2^r = n$, then the induction step is concluded because $f(2^r) = 2^r$. Otherwise let $n = 2^r + s$, where $1 \le s < 2^r$.

We will use $a = n = 2^r + s$, $b = 2^r - s + 1$ with the second condition. Since $2^r - s + 1 \ge 2$ the induction hypothesis implies that $f(2^r - s + 1) = 2^r - s + 1$. So,
$$\begin{align*} f(n) + f(2^r - s + 1) &> f(2^r + s + 2^r - s + 1 - 1) = f(2^{r+1}) \\ \implies f(n) &> f(2^{r+1}) - f(2^r - s + 1) = 2^r + s - 1 = n - 1 \\ \implies f(n) &\ge n. \end{align*}$$
We have shown that $f(n) = n$ which concludes the induction step. All that remains is to determine $f(1)$. The only condition is that $f(1) < 2f(2) = 4$. So, $f(1)$ can be either $1$, $2$ or $3$.

We have obtained the solutions $f(n) = 1$ for all $n$ and $f(n) = n$ for $n \ge 2$ with $f(1) \in \{1, 2, 3\}$. It is easy to check that all of these satisfy the conditions of the problem.