Let us define the position for this game with a pair of numbers $(m, n)$, where $m$ – the number written on the board, and $n$ – the number of a move. We assume a position being winning, if after having got into it, a player still has a winning strategy. A losing one – when having got into it, an opponent will have a winning strategy. Each player tries to reach the position $(0, n)$ after his turn (that is to write down $0$), that is why these positions are winning. Let us show that the map of winning-losing positions looks like in fig. 25.

Fig. 25
Let us note, that from the cell $(m, n)$ one can get into the cell $(m-1, n+1)$, $(m-2, n+1)$, ..., $(m-n, n+1)$ that is, one can move into the next right column, or one, two, ..., $n$ cells down.
Having considered the values $n=1, n=2, n=3$, one can suggest that periods of winning and losing positions in the $n$-th column are as follows:
$$\textit{winning:} [k^2 + nk; k^2 + nk + k], \textit{losing:} [k^2 + nk + k + 1; k^2 + nk + 2k + n], k \in \mathbb{Z}^{+}.$$
We can prove this. That is, show that from each defined as a winning position one can get only into defined as a losing one, and from each position, defined as a losing, there exists a way into a position, defined as a winning.
Let on the $n$-th step, number $m$ is written on the board, and there exists $k \in \mathbb{N}$:
$$k^2 + nk \le m \le k^2 + nk + k \text{ (that is, the position is winning).}$$
Then we can subtract number $l$, which is $1 \le l \le n$. The next move after that will be the $(n+1)$th. We have got:
$$m-l \ge m-n \ge k^2 + nk - n = (k-1)^2 + (n+1)(k-1) + (k-1) + 1,$$
$$m-l \le m-1 \ge k^2 + nk + k-1 = (k-1)^2 + (n+1)(k-1) + 2(k-1) + (n+1).$$
As these are the final points of one losing range, then any move out of the winning position leads into the losing range for $(k-1)$.
Let on the $n$-th step, number $m$ is written on the board and there exists $k \in \mathbb{N}$:
$$k^2 + nk + k + 1 \le m \le k^2 + nk + 2k + n \text{ (that is, the position is losing).}$$
Let us discover what happens if we subtract $1$ or $n$. The move after that will be the $(n+1)$th. We have got:
$$k^2 + (n+1)k + 2k + (n+1) \ge k^2 + nk + n-1 \ge m-1 \ge k^2 + nk + k = k^2 + (n+1)k,$$ that is, we will get into a losing range for $k$, or into a winning position.
$$k^2 + (n+1)k + k = k^2 + nk + 2k \ge m-n \ge k^2 + nk + k + 1 - n \ge$$
$$\ge (k-1)^2 + (n+1)(k-1) + (k-1) + 1,$$
that is, we will get into a losing range for $(k-1)$, or into a winning position.
We can also get from number $m$ to any number from $(m-n)$ to $(m-1)$. As these numbers belong to various losing ranges, there exists a winning range between them.
Thus, the chosen ranges are really winning or losing positions.
Let us see, into which range the position $(2016; 1)$ will get. Just as
$$[44^2 + 44 = 1980 < 2016 < 44^2 + 44 + 44 = 2024,$$
so this position is winning, hence the second player wins.