Given two positive integers $m,n$, we say that a function $f : [0,m] \to \mathbb{R}$ is $(m,n)$-[i]slippery[/i] if it has the following properties:

i) $f$ is continuous;
ii) $f(0) = 0$, $f(m) = n$;
iii) If $t_1, t_2\in [0,m]$ with $t_1 < t_2$ are such that $t_2-t_1\in \mathbb{Z}$ and $f(t_2)-f(t_1)\in\mathbb{Z}$, then $t_2-t_1 \in \{0,m\}$.

Find all the possible values for $m, n$ such that there is a function $f$ that is $(m,n)$-slippery.
1. **Claim**: There exists an \((m,n)\)-slippery function if and only if \(\gcd(m,n) = 1\).

2. **Proof for \(\gcd(m,n) = 1\)**:
   - Consider the function \(f(x) = \frac{n}{m}x\).
   - Verify the properties of \((m,n)\)-slippery:
     - \(f\) is continuous.
     - \(f(0) = 0\) and \(f(m) = \frac{n}{m} \cdot m = n\).
     - For \(t_1, t_2 \in [0, m]\) with \(t_1 < t_2\) such that \(t_2 - t_1 \in \mathbb{Z}\) and \(f(t_2) - f(t_1) \in \mathbb{Z}\):
       \[
       f(t_2) - f(t_1) = \frac{n}{m}(t_2 - t_1)
       \]
       Since \(f(t_2) - f(t_1) \in \mathbb{Z}\), we have:
       \[
       \frac{n}{m}(t_2 - t_1) \in \mathbb{Z}
       \]
       Given \(\gcd(m,n) = 1\), it follows that \(m \mid (t_2 - t_1)\). Since \(t_2 - t_1 \in [0, m]\), the only possible values are \(0\) or \(m\). Thus, \(t_2 - t_1 \in \{0, m\}\).

3. **Proof for \(\gcd(m,n) > 1\)**:
   - Suppose \(d = \gcd(m,n) > 1\).
   - Define \(g(x) = f\left(x + \frac{m}{d}\right) - f(x)\). Note that \(g\) is continuous.
   - Consider the sum:
     \[
     n = f(m) - f(0) = \left(f\left(\frac{m}{d}\right) - f(0)\right) + \left(f\left(\frac{2m}{d}\right) - f\left(\frac{m}{d}\right)\right) + \ldots + \left(f(m) - f\left(\frac{(d-1)m}{d}\right)\right)
     \]
     This can be written as:
     \[
     n = g(0) + g\left(\frac{m}{d}\right) + \ldots + g\left(\frac{(d-1)m}{d}\right)
     \]
   - If any \(g\left(i \frac{m}{d}\right) = \frac{n}{d}\), then \(\frac{n}{d} \in \mathbb{Z}\) and \(m > \frac{m}{d} = t_2 - t_1 \in \mathbb{Z}^+\), which contradicts the condition that \(t_2 - t_1 \in \{0, m\}\).
   - If no summand equals \(\frac{n}{d}\), then there must be one summand greater than \(\frac{n}{d}\) and one smaller. By the Intermediate Value Theorem, there exists \(x\) such that:
     \[
     f\left(x + \frac{m}{d}\right) - f(x) = g(x) = \frac{n}{d}
     \]
     Setting \(t_2 = x + \frac{m}{d}\) and \(t_1 = x\), we obtain a contradiction.

Thus, there is no \((m,n)\)-slippery function if \(\gcd(m,n) > 1\).

The final answer is \( \boxed{ \gcd(m,n) = 1 } \).