5. Let the set $P=\{1,2, \cdots, 2014\}, A \subseteq P$. If any two numbers in set $A$ have a difference that is not a multiple of 99, and the sum of any two numbers is also not a multiple of 99, then the set $A$ can contain at most $\qquad$ elements.
5.50.

Let the set
$$
B_{i}=\{99 \times 1+i, 99 \times 2+i, \cdots, 99 \times 20+i\} \text {, }
$$

where, $i=0,1, \cdots, 34$;
$$
B_{j}=\{99 \times 1+j, 99 \times 2+j, \cdots, 99 \times 19+j\} \text {, }
$$

where, $j=35,36, \cdots, 98$.
Take any $a, b \in A$.
Since the difference between any two numbers in set $A$ is not a multiple of 99, $a$ and $b$ do not belong to the same set $B_{i}$ or $B_{j}$.

Furthermore, since the sum of any two numbers in set $A$ is also not a multiple of 99, $a$ and $b$ are not a pair in the set
$$
M=\left\{\left(B_{1}, B_{98}\right),\left(B_{2}, B_{97}\right), \cdots,\left(B_{49}, B_{50}\right)\right\}
$$

Therefore, the set $A$ can have at most 50 elements, such as taking one element from each of the sets $B_{0}, B_{1}, \cdots, B_{49}$.