We wrote down the sequence of consecutive positive integers starting from 1 on the board, up to a certain number, and then erased one of the written numbers. The arithmetic mean of the remaining numbers is $\frac{602}{17}$. What was the erased number?
I. solution. Let's assume that originally $n$ numbers were written on the board, and the erased number is denoted by $i$. According to the condition,

$$
A=\frac{1+2+\ldots+(n-1)+n-i}{n-1}=\frac{602}{17}=35 \frac{7}{17}
$$

The sum of the first $n-1$ positive integers is $\frac{n(n-1)}{2}$, hence

$$
A=\frac{n}{2}+\frac{n-i}{n-1}
$$

The second term on the right side falls between 0 and 1 (including the boundaries), so the first term must satisfy $347 / 17 \leqq n / 2 \leqq 357 / 17$. Meanwhile, the first term, $n / 2$, is either an integer or half of an integer, so its value is either 34.5 or 35.

If $n / 2=34.5$, then $n=69$, and thus $\frac{69-i}{68}=\frac{7}{17}+\frac{1}{2}$, from which $i=7$. Our steps are reversible, so this case is possible.

If $n / 2=35$, then $n=70$, and thus $\frac{70-i}{69}=\frac{7}{17}$, from which $i=\frac{707}{17}$, which is not an integer, so we do not get a solution in this case.

Therefore, the number 7 was erased from the board.

II. solution. Denoting the sum of the numbers left on the board by $S$,

$$
S=\frac{602}{17} \cdot(n-1)
$$

Since $S$ is an integer and $(602 ; 17)=1$, $n-1$ must be divisible by 17. The erased number lies between 1 and $n$, so

$$
2+3+\ldots+n \geqq S \geqq 1+2+\ldots+(n-1)
$$

or

$$
\frac{(n-1)(n+2)}{2} \geqq \frac{602}{17}(n-1) \geqq \frac{(n-1) n}{2}
$$

Rearranging the inequalities, we get

$$
6914 / 17 \geqq n-1 \geqq 6714 / 17
$$

There is only one number divisible by 17 within the obtained bounds, which is 68. From this, $S=2408$, and the erased number is $(1+2+\cdots+n)-S=2415-2408=7$.

Therefore, the erased number was 7.

Gábor Pál (Budapest, Leőwey K. Secondary School, 8th grade)