5. Let the set $I=\{1,2, \cdots, 2020\}$. We define
$$
\begin{array}{l}
W=\{w(a, b)=(a+b)+a b \mid a, b \in I\} \cap I, \\
Y=\{y(a, b)=(a+b) \cdot a b \mid a, b \in I\} \cap I, \\
X=W \cap Y,
\end{array}
$$

as the "Wu" set, "Yue" set, and "Xizi" set, respectively. Find the sum of the largest and smallest numbers in the Xizi set.
5. If a positive integer $n \in W$, then there exist positive integers $a, b$ such that $n=a+b+a b$.
Thus, $n+1=(a+1)(b+1)$.
Therefore, a positive integer $n \in W$ if and only if $n+1$ is a composite number. Hence,
(1) The set
$\{1,2,4,6,10,12, \cdots\}=\{p-1 \mid p$ is a prime $\}$, any number in this set (referred to as a pre-prime number) does not belong to the set $W$, while all other numbers are in $W$.
(2) For the set $Y, 2=y(1,1) \in Y$, when positive integers $a, b$ are not both 1, $y(a, b)$ is a composite number, i.e., every odd prime does not belong to $Y$. For such numbers, either they are the product of two consecutive positive integers $(y(1, n)=(n+1) n)$, or they can be expressed as the product of three factors, where the largest factor is the sum of the other two $(y(a, b)=(a+b) a b)$.

Notice that, $20=w(2,6)=y(1,4)$, and among the positive integers less than 20, removing the primes and pre-prime numbers, the remaining composite numbers are four, namely $8, 9, 14, 15$, which belong to $W$ but not to $Y$. Therefore, the smallest element of the subset $X$ is 20.
Next, find the largest element of the subset $X$.
By $2000=w(2,666)=y(10,10)$, we know $2000 \in Y$.
Now, we will show that it is the largest number in $X$.
Among the numbers greater than 2000 in $I$, removing the three primes $2003, 2011, 2017$ and their corresponding pre-prime numbers, the remaining 14 numbers can be factored as follows:
$$
\begin{array}{l}
2001=3 \times 23 \times 29,2004=2^{2} \times 3 \times 167, \\
2005=5 \times 401,2006=2 \times 17 \times 59, \\
2007=3^{2} \times 223,2008=2^{3} \times 251, \\
2009=7^{2} \times 41,2012=2^{2} \times 503, \\
2013=3 \times 11 \times 61,2014=2 \times 19 \times 53, \\
2015=5 \times 13 \times 31,2018=2 \times 1009, \\
2019=3 \times 673,2020=2^{2} \times 5 \times 101
\end{array}
$$

Clearly, none of them satisfy the condition in (2). Therefore, they do not belong to the set $Y$.
Thus, the largest number in the subset $X$ is 2000.
Therefore, the sum of the largest and smallest numbers in $X$ is
$$
2000+20=2020 .
$$