Let $f:[0,n]\to \Bbb R$ be continuous with $f(0)=f(n)$. Then there are $n$ pairs of numbers $x,y$ such that $f(x)=f(y)$ and $y-x\in\Bbb N$. 
Theorem. Let $f:[0,n]\to \Bbb R$ be continuous with $f(0)=f(n)$ ($n\in\Bbb N$). Then there exist (at least) $n$ distinct pairs of numbers $x,y$ which satisfy $f(x)=f(y)$ and $y-x\in \mathbb{N}$ (where $0$ is not a natural number).

Partial results (see the two answers below):
Proposition. For $f$ as in the Theorem there exists a $x\in[0,n]$ such that $f(x)=f(x+1)$.
Proof. Define $g(x)=f(x+1)-f(x)$ where $x\in[0,n-1]$. Note that $\sum_{i=0}^{n-1}g(i)=f(n)-f(0)=0$. If all $g(i)=0$ then the proposition holds trivially. Otherwise there must be $i\neq j$ such that $g(i)$ and $g(j)$ have different sign. The proposition now follows from the Intermediate Value Theorem.
Proposition. The Theorem holds under the additional assumption that $f$ is concave or convex.
Proof. See the answer by @Maximilian Janisch.
*Remark.*It is not for each $0<m\leq n$, there must exist $x$ s.t. $f(x)=f(x+m)$. For example, if $f_{[0,1]}(x)>0 \wedge f_{[n-1,n]}(x)<0$, then there doesn't exist $x$ s.t. $f(x)=f(x+n-1)$. However,for some $m$, it may have more than one $x$ satisfying $f(x)=f(x+m)$.

I thought it would be nice to present a direct proof, so pardon me for answering an old question.
First extend $f$ periodically, then for $1\le j<n$, define $g_j(x)=f(x)-f(x+j)$. Notice $g_j$ is $n$-periodic because $f$ is. I will show each $g_j$ has at least 2 roots in $[0,n)$.
First suppose $g_j$ has no roots. Since $g_j$ is continuous, it must be either positive or negative. If it is positive, by periodicity, there is some $c>0$ such that $g_j(x)\ge c$ for all $x$. Then $$(n-1)c \le \sum_{l=1}^{n-1} g_j(x+lj) = f(x) - f(x+nj) = 0,$$ because $f$ is $n$-periodic. This gives a contradiction. Hence, $g_j$ must have one root, call it $x_0$, so all $x_0 + n\mathbb{Z}$ are roots of $g_j$. The case $g_j$ negative is similar.
Next, suppose for contradiction that $g_j$ has only one root in $[0,n)$. Then $g_j$ must be nonnegative or nonpositive by periodicity. Say $g_j$ is nonnegative. By assumption, all roots of $g_j$ are $x_0 + n\mathbb{Z}$, then $$ 0 < \sum_{l=1}^{n-1} g_j(x_0+1/2+lj) = f(x+1/2) - f(x+1/2+nj) = 0,$$ a contradiction.
Finally, notice each root $t$ of $g_j$ gives a pair of points $(t\mod n,\ t+j\mod n)$ satisfying the requirement, and the distance between them is either $j$ or $n-j$. For $1\le j < n/2$, the 2 roots of $g_j$ give distinct pairs. If $n$ is even, we also need 1 root of $g_{n/2}$. These in total give $n-1$ distinct pair of points. Adding $(0,n)$ completes the question.
