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 does he need to try at least to ensure that the safe will definitely open?
Four, 50 times.
First, introduce several notations. Let
$E=\{1,3,5,7,9\}, F=\{0,2,4,6,8\}$
$\Phi=E \cup F=\{0,1, \cdots, 9\}$,
$\Omega=\{x y z \mid x, y, z \in \Phi\}$ (the set of all three-digit codes),
$\Omega_{11}=\{x a b \in \Omega \mid x \in \Phi, a, b \in E\}$,
$\Omega_{12}=\{a y b \in \Omega \mid y \in \Phi, a, b \in E\}$,
$\Omega_{13}=\{a b z \in \Omega \mid z \in \Phi, a, b \in E\}$,
$\Omega_{1}=\Omega_{11} \cup \Omega_{12} \cup \Omega_{13}$ (the set of all three-digit codes with at least two odd numbers),
$\Omega_{2}=\{c d w, c w d, w c d \in \Omega \mid w \in \Phi, c, d \in F\}$ (the set of all three-digit codes with at least two even numbers).
Then $\Omega=\Omega_{1} \cup \Omega_{2}$.
(1) Existence.
Prove that a subset $X$ of $\Omega$ with 50 elements can be constructed such that any three-digit code in $\Omega$ can be opened by a three-digit code in $X$.

Select 25 numbers $x_{a b}, y_{a b}, z_{a b} \in \Phi$, $a, b \in E$, and construct three subsets of $\Omega$ each with 25 elements:
$X_{11}=\left\{x_{a b} a b \in \Omega \mid a, b \in E\right\}$,
$X_{12}=\left\{a y_{a b} b \in \Omega \mid a, b \in E\right\}$,
$X_{13}=\left\{a b z_{a b} \in \Omega \mid a, b \in E\right\}$.
Then for $i=1,2,3$, any three-digit code in the set $\Omega_{1 i}$ can be opened by a three-digit code in the set $X_{1 i}$. Thus, 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:
$X_{1}=X_{11}=X_{12}=X_{13}$
$=\left\{\begin{array}{lllll}111 & 133 & 155 & 177 & 199 \\ 319 & 331 & 353 & 375 & 397 \\ 517 & 539 & 551 & 573 & 595 \\ 715 & 737 & 759 & 771 & 793 \\ 913 & 935 & 957 & 979 & 991\end{array}\right\}$

And any three-digit code in $\Omega_{1}$ can be opened by one of these 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 $|X| = 50$, 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 minimum.
We will prove by contradiction that 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 $|X| \geqslant 50$.
Assume $w=|X| \leqslant 49$, 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.

Without loss of generality, 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, 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, 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, which is $m_{3}$ times. Since at least two sequences contain all the digits $0,1, \cdots, 9$, then
$m \leqslant\left[\frac{49}{10}\right]=4$.
Consider the subset $Y$ of $X$ consisting of elements with the last digit $z$, say
$Y=\left\{\left(a_{1} b_{1} z\right),\left(a_{2} b_{2} z\right), \cdots,\left(a_{m} b_{m} z\right)\right\}$
Let $U=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$,
$V=\left\{b_{1}, b_{2}, \cdots, b_{m}\right\}$

and $U$ contains $s$ different digits, $V$ contains $t$ different digits, then
$1 \leqslant s \leqslant m, 1 \leqslant t \leqslant m$.
Note: $0,1, \cdots, 9$ have $10-s$ digits not appearing in $U$, and $10-t$ digits not appearing in $V$.
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 exists 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
$Z^{\prime}=\{a b c \in X \mid a \notin U, b \notin V, c \neq z\}$.
Then $\left|Z^{\prime}\right| \geqslant(10-s)(10-t)$.
Construct the subset of $X$
$W=\{(a b c) \mid(a b c) \in X, a \in U\}$
Since $U$ contains $s$ different digits, and each digit appears at least $m$ times in $X$, then $|W| \geqslant m s$.
By the construction of sets $Z^{\prime}$ and $W$, it is clear that $W \cap Z^{\prime}=\varnothing$.
Then $49 \geqslant w=|X| \geqslant m s+(10-s)(10-t)$
Thus $49 \geqslant 100-10(s+t)+s t+m s$
$=2(5-s)(5-t)+50+s(m-t) \geqslant 50$,

which is a contradiction.