Example: $10 f(x)$ is a continuous function defined on the interval $[0,2015]$, and $f(0)=f(2015)$. Find the minimum number of real number pairs $(x, y)$ that satisfy the following conditions:
(1) $f(x)=f(y)$;
(2) $x-y \in \mathbf{Z}_{+}$.
$(2015$, Peking University Mathematics Summer Camp)
【Analysis】Consider a continuous function $f(x)$ on the interval $[0, n]$ that satisfies $f(0)=f(n)$. Let the minimum number of real number pairs $(x, y)$ that satisfy (1) and (2) be $F(n)$.
Obviously, $F(1)=1, F(2)=2$.
We will prove: $F(n)=n\left(n \in \mathbf{Z}_{+}\right)$.
Assume that for $n \leqslant k-1$, equation (1) holds.
We need to prove that $F(k)=k$.
First, we prove that $F(k) \geqslant k$.
In fact, take any continuous function $f(x)$ on the interval $[0, k]$ that satisfies $f(0)=f(k)$.
Without loss of generality, assume $f(0)=f(k)=0$.
If there exists $t \in\{1,2, \cdots, k-1\}$ such that $f(t)=0$, then consider the two intervals $[0, t]$ and $[t, k]$.
By the induction hypothesis, it is easy to see that
$$
\begin{array}{l}
F(k) \geqslant F(t)+F(k-t)+1 \\
=t+(k-t)+1=k+1 .
\end{array}
$$

Now assume that $f(1), f(2), \cdots, f(k-1)$ are all non-zero. Without loss of generality, assume $f(1)>0$ (if $f(1)<0$, the proof is similar).
Then
$f(k-1)-f(0)=f(k-1)>0$,
$f(k)-f(1)=-f(1)<0$.
Hence, there exists $t_0 \in (1, k-1)$ such that $f(t_0)=0$.
Since $f(k-1)-f(t_0-1)>0$, there exists $j \in (k-1, k)$ such that $f(j)=f(j+k-t_0)$.

By the induction hypothesis, the number of real number pairs $(x, y)$ that satisfy (1) and (2) in the interval $\left[j, j+k-t_0\right]$ is at least $k-t_0$.
In the interval $\left[0, t_0\right]$,
$f(0)=0, f(1)>0, f(2)>0$,
$\cdots, f(t_0-1)>0, f(t_0)<0, f(t_0)-f(t_0-m)<0$.
Then there exist $x, y \in \left[0, t_0\right]$ such that
$f(x)=f(y)$, and $x-y=m$.
Additionally, $(x, y)=(2015,0)$ is also a pair that satisfies (1) and (2).
Therefore, $F(k) \geqslant (k-t_0) + (t_0-1) + 1 = k$.
In summary, $F(k) \geqslant k$.
Next, we provide an example to show that $F(k)=k$.
In fact, any function $f(x)$ that first increases and then decreases satisfies $F(k)=k$.

This is because, for any positive integer $m \leqslant k$, the number of pairs $(x, y)$ that satisfy $f(x)=f(y)$ and $x-y=m$ is exactly one.
Thus, $F(k)=k$, i.e., for any $n \in \mathbf{Z}_{+}$, we have
$F(n)=n$.
Therefore, $F(2015)=2015$.