Solution:

It is immediate to verify that functions of the type $f(n) = (n - n_{0}) a$ with $n_{0}$ an integer and $a$ a positive real number satisfy the hypotheses: if $m < n$ then $(m - n_{0}) a < (n - n_{0}) a$ and $f(m) - f(n) = (m - n_{0}) a - (n - n_{0}) a = m a - n a = [(m - n + n_{0}) - n_{0}] a = f(k)$ with $k = m - n + n_{0}$.

We show that these functions are the only possible ones.

Let $f$ be a function satisfying the given conditions. Setting $m = n$ in the second condition we obtain that there exists an integer $n_{0}$ such that $0 = f(n) - f(n) = f(n_{0})$.

Let then $n_{0}$ be an integer such that $f(n_{0}) = 0$ and set $a = f(n_{0} + 1)$; we prove by induction that the real numbers of the form $k a$ with $k$ an integer are values of the function $f$. If $n a = f(m)$ is a value, from $f(n_{0}) - f(m) = -n a$ we obtain that $-n a$ is also a value; suppose that $n a$ is a value: then also $-a - n a = -(n + 1) a$ and $n a - (-a) = (n + 1) a$ are values. Hence $f$ must take as values all integer multiples of $a$.

We now prove that only numbers of this form are values of $f$. Let $b$ be a real number such that $b / a$ is not an integer (note that $a > f(n_{0}) = 0$ by the first property, so we can always divide by $a$), and suppose for contradiction that $b$ is a value of $f$. If $b > 0$, consider the largest natural number $k$ such that $k a$ is less than $b$; since we have already seen that $k a$ is attained as a value by $f$, this must also happen for $b - k a$. But by our choice of $k$ we have $k a < b < (k + 1) a$, hence $0 < b - k a < a$, and this is a contradiction: the function is strictly increasing, hence does not take values between $f(n_{0}) = 0$ and $f(n_{0} + 1) = a$. The same reasoning holds for $b < 0$: letting $k$ be the smallest natural number such that $-k a < b$, the number $b - (-k a)$, lying between $0$ and $a$, should belong to the image. Again this is a contradiction.

We have thus shown that the possible functions are all and only the increasing ones whose image consists of the integer multiples of a positive real number $a$, that is, those of the form: $f(n_{0}) = 0$ for some integer $n_{0}$; $f(n_{0} + k) = k a$ for every integer $k$, for some positive real $a$.

Solution:

As in the previous solution, we note that all functions of the type $f(n) = (n - n_{0}) a$ with $n_{0}$ an integer and $a$ a positive real number satisfy the given conditions.

Again analogously to before, let $n_{0}$ be an integer such that $f(n_{0}) = 0$ and let $a = f(n_{0} + 1)$ ($a$ is positive by the first condition). We must show that necessarily $f(n) = (n - n_{0}) a$ for every $n$ or, equivalently, that $g(n) = f(n + n_{0}) = n a$ for every $n$. It is immediate to verify that the function $g$ satisfies the same conditions given for $f$.

Let $n$ be any integer. Since the function $g$ is strictly increasing, we have $g(n + 1) - g(n) > 0$, and hence, since $g(n + 1) - g(n)$ is a value of the function, $g(n + 1) - g(n) \geq g(1)$; on the other hand, $g(n + 1) - g(1) < g(n + 1)$ and hence, since $g(n + 1) - g(n)$ is a value of the function, $g(n + 1) - g(1) \leq g(n)$. In conclusion,
$$g(n + 1) = g(n) + g(1) \quad \text{for every integer } n.$$
Now we prove by induction that $g(n) = n a$ for every integer $n$. We know that this is true for $n = 0, 1$. Suppose that the claim is true for some positive integer $n$: then the previous formula gives $g(n + 1) = n a + a = (n + 1) a$. Suppose now that the claim is true for some $m \leq 0$: setting $n = m - 1$ in the previous formula we obtain $g(m) = g(m - 1) + g(1)$, from which $g(m - 1) = g(m) - g(1) = m a - a = (m - 1) a$.