11.5. In Flower City, there live $99^{2}$ dwarfs. Some of the dwarfs are knights (always tell the truth), while the rest are liars (always lie). The houses in the city are located in the cells of a $99 \times 99$ square (a total of $99^{2}$ houses, arranged in 99 vertical and 99 horizontal streets). Each house is inhabited by exactly one dwarf. The house number is denoted by a pair of numbers $(x ; y)$, where $1 \leqslant x \leqslant 99$ is the number of the vertical street (numbers increase from left to right), and $1 \leqslant y \leqslant 99$ is the number of the horizontal street (numbers increase from bottom to top). The Flower City distance between two houses with numbers $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$ is the number $\rho=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right|$. It is known that on each street, whether vertical or horizontal, there live at least $k$ knights. In addition, all dwarfs know which house Knight Knowitall lives in. You want to find his house, but you do not know what Knowitall looks like. You can approach any house and ask the dwarf living there: “What is the Flower City distance from your house to the house of Knowitall?”. For what smallest $k$ can you guarantee to find the house of Knowitall? (V. Novikov)
Answer: 75.

Solution: Example. Let's show that if $k=74$, we cannot guarantee finding the house of Znayka. Place Znayka and the liar Neznayka in houses with numbers $(50 ; 49)$ and $(49 ; 50)$, respectively. We will show that it might be such that from the answers of the residents, we cannot uniquely determine which of these two houses Znayka lives in.

In the lower left square $49 \times 49$, we will settle the knights. Their distances to Znayka and Neznayka are the same. In the upper right square $50 \times 50$, we will also settle the knights, and their distances are also the same. In the lower right rectangle (49 rows and 50 columns), we will settle the knights so that in each row there are exactly 25 knights and 25 liars, and in each column at least 24 knights and 24 liars. In the upper left rectangle, we will settle the dwarfs diagonally symmetric to the upper right, but swap the knights and liars. In it, each column will have 25 knights and 25 liars, and each row will have at least 24 knights and 24 liars. For each dwarf in these rectangles, the distances to Znayka and Neznayka are different. Let all the liars in them say the distance not to Znayka, but to Neznayka. Then, when the places of the knights and liars in these rectangles are swapped (in particular, when Znayka and Neznayka are swapped), everyone will say the same thing, but Znayka will live in a different house.

Estimate. Let's show that if $k \geqslant 75$, we can guarantee finding the house of Znayka. Suppose that, after asking all the dwarfs, we cannot determine where Znayka is, i.e., there are at least two houses where he could be. Let one have the number $(x ; y)$, and the other $(u ; v)$. We can assume that $x \leqslant u, y \leqslant v$, since we can rotate the square as required. Since both inequalities cannot be equalities at the same time, without loss of generality, we will assume that $yv-y$ or $(u-x)+(v-y)$ is odd, then in these columns there is no dwarf whose distance to both marked houses is the same.

If $x=u$ and $v-y$ is even, then in the column $(x, \ldots)$ there is one dwarf whose distance to both houses is the same.

If $u-x<v-y$ and $(u-x)+(v-y)$ is even, then in the columns $(x, \ldots)$ and $(u, \ldots)$ there is one dwarf in each whose distance to both houses is the same.

If $u-x=v-y$, then in the column $(x, \ldots)$, the places from which the distances to $(x ; y)$ and $(u ; v)$ are the same have the form $(x ; V)$, where $V \geqslant v$, and there are exactly $100-v$ such places. Similarly, in the column $(u, \ldots)$, the places from which the distances to $(x ; y)$ and $(u ; v)$ are the same have the form $(u ; Y)$, where $Y \leqslant y$, and there are exactly $y$ such places.

Notice that $y+(100-v) \leqslant 99$. Therefore, one of the numbers $y, 100-v$ is not greater than 49.

Thus, in all cases, there will be a column where no more than 49 knights indicate both places, while on the incorrect place, no more than these knights and all the liars (no more than $99-75=24$) point, i.e., no more than $49+24=73$ dwarfs. At the same time, on the correct place, in any column, at least all the knights point, i.e., no fewer than 75 dwarfs. Thus, from the two suspicious places, one can always be excluded (since the row or column we rely on depends only on the position of the places, not on the arrangement of the knights/liars). This means that the correct place can always be found.

Comment: The correct answer without explanation - 0 points. Only the example - 3 points.

Only the estimate - 4 points.