116. The distance from point $A$ to point $B$ is 899 km. Along the road connecting $A$ and $B$, there are kilometer markers indicating the distances to $A$ and to $B$ - the inscriptions on them look like this:

$$
0|999,1| 998,2|997, \ldots, 999| 0
$$

How many of these markers have only two different digits?
116. It is clear that if the numbers on the pole are $\overline{\lambda y z} \mid \overline{x_{1} y_{1} z_{i}}$ (where $x$, $\boldsymbol{y}, \ldots$ are digits), then $\overline{x_{1} y_{1} z_{1}}=999-x y z$ and, therefore, $z_{1}=9-z, y_{1}=$ $=9-y, x_{1}=9-x$. (If $x=9$ or $x=y=9$, then the digits $x_{1}=$ $=0$, or $x_{1}=y_{1}=0$ are not indicated on the pole.) From this, it immediately follows that if $x=y=z$ (and, therefore, $x_{1}=y_{1}=z_{1}=9-x$), then our conditions will be satisfied; this gives 10 poles satisfying the condition of the problem (corresponding to distances $0=000$, 11, $222, \ldots, 999$ km from point $A$). If the number $\overline{x y z}$ is written with two different digits, then these digits must be such that each of them complements the other to the number 9—in this case, the number $\overline{x_{1} y_{1} z_{1}}=9-x, 9-y \cdot 9-z$ will also be written with the same digits. Clearly, there are five such pairs of digits: $(0,9),(1,8),(2,7)$, $(3,6)$, and $(1,5)$; and there are six three-digit numbers written with two given digits $a$ and $b$: three of them are written with two digits $a$ and one digit $b$ (which can occupy any of the 3 available positions) and, similarly, three numbers are written with one digit $a$ and two digits $b$. Thus, we get another $5 \cdot 6=30$ distances satisfying the condition of the problem for the pole from point $A$; therefore, the total number of poles satisfying the formulated requirements is: $10+30=40$.