Problem 3. Several (more than one) consecutive natural numbers are written on the board, the sum of which is 2016. What can the smallest of these numbers be?
Answer: $1,86,220,285$ or 671.

Solution. Suppose that for some $n$ there exist $n$ consecutive natural numbers $a$, $a+1, \ldots, a+(n-1)$, the sum of which is 2016. Then $n a+(n-1) n / 2=2016$, or, after algebraic transformations, $n(2 a+n-1)=4032=2^{6} \cdot 3^{2} \cdot 7$.

Note that $n$ and $2 a+n-1$ have different parity. Therefore, if $n$ is even, then $n$ must be divisible by $2^{6}=64$, hence $n \geqslant 64,2 a+n-1 \leqslant 4032 / 64=63$, which contradicts $2 a+n-1>n$.

Thus, $n$ is an odd divisor of the number 4032, i.e., a divisor of the number 63. Let's check that each of them fits and find the corresponding values of $a$. If $n=3: 2 a+3-1=1344, a=671$; if $n=7$ : $2 a+7-1=576, a=285$; if $n=9: 2 a+9-1=448, a=220$; if $n=21: 2 a+21-1=192, a=86$; if $n=63: 2 a+63-1=64, a=1$.