4. On a sheet of paper, 12 consecutive integers are written. After one of them is crossed out, the sum of the remaining numbers equals 325. Which number was crossed out?
Solution. Let $n, n+1, \ldots, n+k-1, n+k, n+k+1, \ldots, n+11$ be 12 consecutive integers, and the number $n+k, k=0,1,2, \ldots, 11$ is crossed out. The sum of the numbers after crossing out is

$$
\frac{2 n+11}{2} \cdot 12-(n+k)=325 \rightarrow 11 n+66-k=325 \rightarrow k=11 n-259
$$

Considering the condition $k \in[0 ; 11]$, we get $0 \leq 11 n-259 \leq 11 \rightarrow \frac{259}{11} \leq n \leq \frac{270}{11}$. In this interval, there is only one integer $n=24$ and $k=5$. Therefore, the crossed-out number is $n+k=29$.

Answer: 29.