10. (20 points) Given the sequence $\left\{a_{n}\right\}(n \in \mathbf{N})$ satisfies: $a_{1}=1$, and for any non-negative integers $m, n (m \geqslant n)$, we have
$$
a_{m+n}+a_{m-n}+m-n-1=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right) \text {. }
$$

Find the value of $\left[\frac{a_{2013}}{2012}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$).
10. Let $m=n$, we get $a_{0}=1$.

Let $n=0$, we get $a_{2 m}=4 a_{m}+2 m-3$.
In particular, $a_{2}=3$.
Let $n=1$, we get
$$
\begin{array}{l}
a_{m+1}+a_{m-1}+m-2=\frac{1}{2}\left(a_{2 m}+a_{2}\right)=2 a_{m}+m \\
\Rightarrow a_{m+1}-a_{m}=a_{m}-a_{m-1}+2 .
\end{array}
$$

Thus, $a_{m+1}-a_{m}=2 m(m \geqslant 1)$,
$$
\begin{array}{l}
a_{m}=a_{1}+\sum_{k=1}^{m-1}\left(a_{k+1}-a_{k}\right)=1+\sum_{k=1}^{m-1} 2 k \\
=m(m-1)+1(m \geqslant 1) .
\end{array}
$$

Then $\left[\frac{a_{2013}}{2012}\right]=2013$.