One morning at school, consecutive integers starting from 1 were written on the board up to a certain number. The troublemaker erased one of the numbers carefully. The whole incident would have been forgotten if someone hadn't noted that the arithmetic mean of the remaining numbers was $\frac{45}{4}$. Let's try to figure out which number the troublemaker erased.
The numbers written on the board:

$$
1, \quad 2, \quad \ldots, \quad n-1, \quad n
$$

Let the erased number be denoted by $x$. The sum of the first $n$ natural numbers, $S_{n}$, is given by the well-known formula $S_{n}=\frac{n(n+1)}{2}$. Subtract the erased number from the sum, and write the arithmetic mean of the remaining numbers, which is known to be $\frac{45}{4}$; we get

$$
\frac{\frac{n(n+1)}{2}-x}{n-1}=\frac{45}{4}
$$

The erased number is at least 1 and at most $n$, so

$$
\frac{\frac{n(n+1)}{2}-n}{n-1} \leq \frac{45}{4} \leq \frac{\frac{n(n+1)}{2}-1}{n-1}
$$

Rearranging the inequality,

$$
\frac{n}{2} \leq \frac{45}{4} \leq \frac{n+2}{2}
$$

from which $20.5 \leq n \leq 22.5$. Since $n$ can only be an integer, therefore $n=21$ or 22.

Substituting the obtained values of $n$ into (1):

For $n=21$, $\frac{21 \cdot 11-x}{20}=\frac{45}{4}$, from which $x=6$

For $n=22$, $\frac{11 \cdot 23-x}{21}=\frac{45}{4}$, from which $4 x=67$, $x$ is not an integer, so it is not a solution.

Therefore, the erased number is 6.

Bérces Márton (Budapest, Arany J. Gymnasium, 11th grade)