Four, (50 points) Mr. Wanda often forgets the numbers he should remember, such as his friends' phone numbers, the password of the safe, etc. For this reason, the manufacturer specially designed a password lock for his office safe, with keys labeled $0 \sim 9$. It is known that the password of the safe is a three-digit number, but as long as two of the digits in the entered three-digit number are correct, the safe will open. Mr. Wanda has forgotten the password he set again. How many times at least does he need to try to ensure that the safe will definitely open?
Each code can be opened by a three-digit code in the set $X_{1 i}$. Therefore, any three-digit code in the set $\Omega_{1}$ can be opened by a three-digit code in the set $X_{1}=X_{11} \cup X_{12} \cup X_{13}$.

Next, take special $X_{11}, X_{12}, X_{13}$, such that $X_{11} = X_{12} = X_{13}$, thus, $X_{1} = X_{11} = X_{12} = X_{13}$ is a set containing 25 elements, for example:
$$
\begin{array}{l}
X_{1} = X_{11} = X_{12} = X_{13} \\
= \left\{\begin{array}{lllll}
111 & 133 & 155 & 177 & 199 \\
319 & 331 & 353 & 375 & 397 \\
517 & 539 & 551 & 573 & 595 \\
715 & 737 & 759 & 771 & 793 \\
913 & 935 & 957 & 979 & 991
\end{array}\right\},
\end{array}
$$

and any three-digit code in $\Omega_{1}$ can be opened by one of the 25 three-digit codes in $X_{1}$.

Similarly, a set $X_{2}$ of 25 three-digit codes composed entirely of even numbers can be constructed, such that any three-digit code in $\Omega_{2}$ can be opened by a three-digit code in $X_{2}$.

Let $X = X_{1} \cup X_{2}$. By the construction of $X_{1}$ and $X_{2}$, we know that $|X| = 50$, and any three-digit code in $\Omega = \Omega_{1} \cup \Omega_{2}$ can be opened by a three-digit code in $X$.
(2) 50 is the smallest.
We will prove by contradiction: If a subset $X$ of $\Omega$ satisfies that any three-digit code in $\Omega$ can be opened by a three-digit code in $X$, then $|X| \geq 50$.
Assume $w = |X| \leq 49$, and let
$X = \left\{\left(a_{1} b_{1} c_{1}\right),\left(a_{2} b_{2} c_{2}\right), \cdots,\left(a_{w} b_{w} c_{w}\right)\right\}$.
Consider the three sequences
$a_{1}, a_{2}, \cdots, a_{w} ; b_{1}, b_{2}, \cdots, b_{w} ; c_{1}, c_{2}, \cdots, c_{w}$.
Then, in these three sequences, at least two sequences must contain all the digits $0,1, \cdots, 9$. Otherwise, suppose the digit $a (a \in \Phi)$ does not appear in $a_{1}, a_{2}, \cdots, a_{w}$, and the digit $b (b \in \Phi)$ does not appear in $b_{1}, b_{2}, \cdots, b_{w}$, then the three-digit code $a b 0 \in \Omega$ cannot be opened by any three-digit code in $X$, which is a contradiction.

Assume the set of different digits appearing in $a_{1}, a_{2}, \cdots, a_{w}$ is $A$, and let $x \in A$ appear the fewest times in $a_{1}, a_{2}, \cdots, a_{w}$, which is $m_{1}$ times.

Similarly, the set of different digits appearing in $b_{1}, b_{2}, \cdots, b_{w}$ is $B$, and let $y \in B$ appear the fewest times in $b_{1}, b_{2}, \cdots, b_{w}$, which is $m_{2}$ times; the set of different digits appearing in $c_{1}, c_{2}, \cdots, c_{w}$ is $C$, and let $z \in C$ appear the fewest times in $c_{1}, c_{2}, \cdots, c_{w}$, which is $m_{3}$ times.

Let $m = \min \left\{m_{1}, m_{2}, m_{3}\right\}$. Assume $m = m_{3}$, since at least two sequences contain all the digits $0,1, \cdots, 9$, then
$$
m \leq \left[\frac{49}{10}\right] = 4.
$$

Consider the subset $Y$ of elements in $X$ with the last digit $z$, let
$$
\begin{array}{l}
Y = \left\{\left(a_{1} b_{1} z\right),\left(a_{2} b_{2} z\right), \cdots,\left(a_{m} b_{m} z\right)\right\}. \\
\text{Let } U = \left\{a_{1}, a_{2}, \cdots, a_{m}\right\}, \\
V = \left\{b_{1}, b_{2}, \cdots, b_{m}\right\},
\end{array}
$$

and $U$ contains $s$ different digits, $V$ contains $t$ different digits, then
$1 \leq s \leq m, 1 \leq t \leq m$.
Note: The digits $0,1, \cdots, 9$ that do not appear in $U$ are $10-s$ in number, and those that do not appear in $V$ are $10-t$ in number.
Construct the set
$Z = \{a b z \in \Omega \mid a \notin U, b \notin V\}$.
Therefore, $|Z| = (10-s)(10-t)$. For each $a b z \in Z$, there is at least one three-digit code $\alpha \beta \gamma \in X$ that can open $a b z$.
By the construction of $Z$, we know $\alpha = a, \beta = b, \gamma \neq z$. Let
$Z' = \{a b c \in X \mid a \notin U, b \notin V, c \neq z\}$.
Then $|Z'| \geq (10-s)(10-t)$.
Construct the subset of $X$
$W = \{(a b c) \mid (a b c) \in X, a \in U\}$.
Since $U$ contains $s$ different digits, each digit appears at least $m$ times in $X$, thus, $|W| \geq m s$.
By the construction of sets $Z'$ and $W$, it is clear that $W \cap Z' = \varnothing$.
Then $49 \geq w = |X| \geq m s + (10-s)(10-t)$.
Hence $49 \geq 100 - 10(s+t) + st + ms$
$$
= 2(5-s)(5-t) + 50 + s(m-t) \geq 50,
$$

which is a contradiction.