7. Sets $A$ and $B$ are defined as follows:
$$
\begin{aligned}
A= & \left\{a^{3}+b^{3}+c^{3}-3 a b c \mid a, b, c \in \mathbf{N}\right\}, \\
B= & \{(a+b-c)(b+c-a)(c+a-b) \mid \\
& a, b, c \in \mathbf{N}\} .
\end{aligned}
$$

Let set $P=\{n \mid n \in A \cap B, 1 \leqslant n \leqslant 2016\}$. Find the number of elements in $P$. (Wu Quanshui, Tao Pingsheng, problem contributor)
7. First, prove two lemmas.

Lemma 1 If $n$ is a positive integer, then $n \in A \Leftrightarrow n \neq 3,6(\bmod 9)$.
Proof of Lemma 1 Let
$$
U(a, b, c)=a^{3}+b^{3}+c^{3}-3 a b c(a, b, c \in \mathbf{N}) \text {. }
$$

For each natural number $k$, we have
$$
\begin{array}{l}
U(k, k, k+1)=2 k^{3}+(k+1)^{3}-3 k^{2}(k+1) \\
=3 k+1 \in A, \\
U(k, k+1, k+1) \\
=k^{3}+2(k+1)^{3}-3 k(k+1)^{2} \\
=3 k+2 \in A, \\
U(k, k+1, k+2) \\
=k^{3}+(k+1)^{3}+(k+2)^{3}-3 k(k+1)(k+2) \\
=9(k+1) \in A .
\end{array}
$$

Thus, when a positive integer $n$ satisfies $n \neq 3,6(\bmod 9)$, it must be that $n \in A$.
Conversely, note that
$$
U(a, b, c)
$$
$$
=(a+b+c)\left[(a+b+c)^{2}-3(a b+b c+c a)\right] \text {. }
$$

When $3 \mid (a+b+c)$, $9 \mid U(a, b, c)$;
When $3 \nmid (a+b+c)$, $3 \nmid U(a, b, c)$.
Therefore, there do not exist natural numbers $a, b, c$ such that
$$
U(a, b, c) \equiv 3,6(\bmod 9) \text {. }
$$

In other words, for any $n \in A$, it must be that $n \neq 3,6(\bmod 9)$.

Lemma 2 If $n$ is a positive integer, then
$$
n \in B \Leftrightarrow n \equiv 0,1,3,5,7(\bmod 8) \text {. }
$$

Proof of Lemma 2 Let
$$
V(a, b, c)=(a+b-c)(b+c-a)(c+a-b),
$$

where $a, b, c \in \mathbf{N}$.
For each natural number $k$,
$$
\begin{array}{l}
V(k+1, k+1,1)=(2 k+1) \times 1 \times 1 \\
=2 k+1 \in B, \\
V(k+1, k+1,2)=2 k \times 2 \times 2=8 k \in B .
\end{array}
$$

Thus, when a positive integer $n$ satisfies
$$
n \equiv 0,1,3,5,7(\bmod 8)
$$

it must be that $n \in B$.
Conversely, note that $a+b-c, b+c-a, c+a-b$ have the same parity.
When they are all even, $8 \mid V(a, b, c)$;
When they are all odd, $2 \mid V(a, b, c)$.
Therefore, there do not exist natural numbers $a, b, c$ such that
$V(a, b, c) \equiv 2,4,6(\bmod 8)$.
In other words, for any $n \in B$, it must be that
$n \neq 2,4,6(\bmod 8)$.
Lemmas 1 and 2 are proved.
Next, find the number of elements in set $P$.
According to Lemmas 1 and 2, we know
$$
\begin{aligned}
P= & \{n \mid 1 \leqslant n \leqslant 2016, n \neq 0,1,3,5,7(\bmod 8), \\
& n \neq 3,6(\bmod 9)\} .
\end{aligned}
$$

First, consider the numbers in $\{1,2, \cdots, 72\}$ that are congruent to $0, 1, 3, 5, 7 \pmod{8}$:
$$
\begin{array}{l}
1, 9, 17, 25, 33, 41, 49, 57, 65, \\
3, 11, 19, 27, 35, 43, 51, 59, 67, \\
5, 13, 21, 29, 37, 45, 53, 61, 69, \\
7, 15, 23, 31, 39, 47, 55, 63, 71, \\
8, 16, 24, 32, 40, 48, 56, 64, 72 .
\end{array}
$$

In the above five rows of numbers, each row forms a complete residue system modulo 9. Therefore, among all the numbers, exactly $5 \times 7=35$ numbers have a remainder that is not $3, 6$ modulo 9, and they belong to $P$.
Each $n \in P$ can be uniquely represented as
$$
n=72 l+m(l \in\{0,1, \cdots, 27\}, m \in\{1,2, \cdots, 72\})
$$

form, then $n \in P$ if and only if $m \in P$.
Therefore, the number of elements in set $P$ is $35 \times 28=980$.