1. Nikola thought of 4 different real numbers and wrote down the product of each pair of them on the board (thus 6 numbers appeared on the board). Dule erased one of those numbers, after which 5 consecutive natural numbers remained on the board. Which number did Dule erase?
1. Let the numbers left on the board be $n, n+1, n+2, n+3$ and $n+4$. Among these numbers, there are two groups of two numbers each such that the product of the numbers in the first group is equal to the product of the numbers in the second group: indeed, if Nikola thought of the numbers $a, b, c$ and $d$, and if Dule erased, without loss of generality, the number $c d$, then we have $(b c)(a d)=(a c)(b d)$ and the numbers $b c, a d, a c$ and $b d$ are on the board. Moreover, the number on the board that does not participate in either of these two groups, when multiplied by the erased number, gives the same product (indeed, the number not used in the previous consideration is $a b$, and when multiplied by the erased number, $c d$, it gives the product $a b c d$ again).

We have five possibilities for the number that does not participate in either of these two groups. Note that, after choosing this number, for the product of any two of the remaining four numbers to be equal to the product of the other two, one group must consist of the smallest and the largest number, and the other group must consist of the two middle numbers (otherwise, the product involving the smallest number would be smaller than the product involving the largest). Therefore, one of the following equalities must hold: $(n+1)(n+4)=$ $(n+2)(n+3), n(n+4)=(n+2)(n+3), n(n+4)=(n+1)(n+3), n(n+4)=(n+1)(n+2)$ or $n(n+3)=(n+1)(n+2)$. These are equivalent to, respectively, $5 n+4=5 n+6$, $4 n=5 n+6,4 n=4 n+3,4 n=3 n+2$ and $3 n=3 n+2$, and we observe that the only one of them that has a solution in the set of natural numbers is the second to last, whose solution is $n=2$.

Thus, the numbers left on the board are $2,3,4,5$ and 6. Among these numbers, there is exactly one way to divide them into two groups of two numbers each such that the products of the numbers in the groups are equal: $2 \cdot 6=3 \cdot 4$. Therefore, the product of the unused number (which is 5) and the erased number must also be 12, so the erased number is $\frac{12}{5}$.