Let the four-digit number be $abcd$, where $a$, $b$, $c$, $d$ are its digits and $a \neq 0$.

A four-digit mountain-shaped number must satisfy:
- The digits first increase, then decrease.
- That is, there exists an index $k$ ($2 \leq k \leq 4$) such that $a < b < \cdots < x$ (up to position $k$), then $x > y > \cdots$ (from position $k$ onward).

But since there are only four digits, the only possible way is:
- $a < b < c > d$ (the peak is at $c$), or
- $a < b > c > d$ (the peak is at $b$).

But in the problem, 'first increase then decrease' means the digits strictly increase, then strictly decrease, and the peak is not at the first or last digit.

So, the only possible form is $a < b < c > d$.

Let us count the number of such numbers.

Let $a$, $b$, $c$, $d$ be digits such that $a < b < c$ and $c > d$.

Since $a$ is the first digit, $a \geq 1$.

Let us fix $a < b < c$ first. The digits $a$, $b$, $c$ are distinct and increasing, and $a \geq 1$, $b$, $c$ are digits from $0$ to $9$.

Let us choose $a$, $b$, $c$ such that $a < b < c$ and $a \geq 1$.

The number of ways to choose $a$, $b$, $c$ with $a < b < c$, $a \geq 1$ is $\binom{9}{3}$, since $a$ can be $1$ to $7$, $b$ and $c$ are larger digits, and all three are distinct.

For each such triple, $d$ is a digit such that $d < c$ (since $c > d$), and $d$ can be any digit from $0$ to $9$, except that $d$ can be equal to $a$ or $b$ (since otherwise the digits would not first increase then decrease), but the problem does not specify that all digits must be distinct, only that the digits first increase then decrease.

But let's check: For $a < b < c > d$, is it allowed for $d$ to be equal to $a$ or $b$? Let's look at the example: $1310$ is mountain-shaped, and $1$ repeats. So, digits can repeat, except that the sequence must first strictly increase, then strictly decrease.

So, $a < b < c$ (strictly increasing), $c > d$ (strictly decreasing from $c$ to $d$), but $d$ can be equal to $a$ or $b$.

So, for each choice of $a < b < c$, $d$ can be any digit from $0$ to $c-1$ (since $d < c$), including possibly $a$ or $b$.

Therefore, for each $a < b < c$, the number of possible $d$ is $c$ (since $d = 0, 1, \ldots, c-1$).

So, the total number is:

$$\sum_{a=1}^{7} \sum_{b=a+1}^{8} \sum_{c=b+1}^{9} c$$

Let us compute this sum.

Let us fix $a$, $b$, $c$ with $a < b < c$, $a \geq 1$, $c \leq 9$.

For $a$ from $1$ to $7$:
- $b$ from $a+1$ to $8$
- $c$ from $b+1$ to $9$

Let us compute the sum:

For each $a$, $b$, $c$, add $c$.

Let us fix $a$ and $b$.
For each $c$ from $b+1$ to $9$, sum $c$.

The sum $S = \sum_{a=1}^{7} \sum_{b=a+1}^{8} \sum_{c=b+1}^{9} c$

Let us compute for each $a$:

For $a=1$:
- $b$ from $2$ to $8$
- For $b=2$: $c$ from $3$ to $9$ (i.e., $3,4,5,6,7,8,9$)
- For $b=3$: $c$ from $4$ to $9$ ($4,5,6,7,8,9$)
- For $b=4$: $c$ from $5$ to $9$ ($5,6,7,8,9$)
- For $b=5$: $c$ from $6$ to $9$ ($6,7,8,9$)
- For $b=6$: $c$ from $7$ to $9$ ($7,8,9$)
- For $b=7$: $c$ from $8$ to $9$ ($8,9$)
- For $b=8$: $c=9$

Let us sum $c$ for each $b$:
- $b=2$: $3+4+5+6+7+8+9=42$
- $b=3$: $4+5+6+7+8+9=39$
- $b=4$: $5+6+7+8+9=35$
- $b=5$: $6+7+8+9=30$
- $b=6$: $7+8+9=24$
- $b=7$: $8+9=17$
- $b=8$: $9$

Sum: $42+39+35+30+24+17+9=196$

Now, $a=2$:
- $b$ from $3$ to $8$
- $b=3$: $c=4$ to $9$: $4+5+6+7+8+9=39$
- $b=4$: $5+6+7+8+9=35$
- $b=5$: $6+7+8+9=30$
- $b=6$: $7+8+9=24$
- $b=7$: $8+9=17$
- $b=8$: $9$

Sum: $39+35+30+24+17+9=154$

$a=3$:
- $b=4$: $5+6+7+8+9=35$
- $b=5$: $6+7+8+9=30$
- $b=6$: $7+8+9=24$
- $b=7$: $8+9=17$
- $b=8$: $9$

Sum: $35+30+24+17+9=115$

$a=4$:
- $b=5$: $6+7+8+9=30$
- $b=6$: $7+8+9=24$
- $b=7$: $8+9=17$
- $b=8$: $9$

Sum: $30+24+17+9=80$

$a=5$:
- $b=6$: $7+8+9=24$
- $b=7$: $8+9=17$
- $b=8$: $9$

Sum: $24+17+9=50$

$a=6$:
- $b=7$: $8+9=17$
- $b=8$: $9$

Sum: $17+9=26$

$a=7$:
- $b=8$: $c=9$

Sum: $9$

Now, sum all these:
$196+154=350$
$350+115=465$
$465+80=545$
$545+50=595$
$595+26=621$
$621+9=630$

Therefore, the total number of four-digit mountain-shaped numbers is $630$.