5. The game starts with the number 2. On a turn, it is allowed to add to the current number any natural number less than it. The player who reaches the number 1000 wins. Who will win with correct play?
5. Let's call a winning position the one from which, by making a move, we win; and a losing position the one from which we lose. Obviously, all positions from which we can move to a winning position are losing (if we move to such a position, the opponent can move to a winning position on the next move and win). Conversely, positions from which we can only move to losing positions are winning (since by making a move to such a position, we force the opponent to move to a losing position). Let's return to our game now. By receiving the number 1000, we win, so 1000 is a winning position. We can reach 1000 from $501, 502, 503, \ldots, 999$, so all these positions are losing. From the number 500, we can only move to one of the numbers $501, 502, 503, \ldots, 999$, that is, only to a losing position. Therefore, 500 is a winning position. Similarly, 251, 252, ... 499 are losing positions, and 250 is a winning position. Continuing this reasoning, we get the winning positions $125, 62, 31, 15, 7, 3$. Thus, the first player's first move lands in a winning position: $2+1=3$. To win, he must follow this strategy: each of his moves should be to a winning position. Clearly, he always has this option. Let's consider this strategy:

| Move | First Player | Second Player |
| :---: | :---: | :---: |
| 1 | $2+1=3$ | $3+a$ |
| 2 | $3+a+4-a=7$ | $7+b$ |
| 3 | $7+b+8-b=15$ | $15+c$ |
| 4 | $15+c+16-c=31$ | $31+d$ |
| 5 | $31+d+31-d=62$ | $62+e$ |
| 6 | $62+e+63-e=125$ | $125+f$ |
| 7 | $125+f+125-f=250$ | $250+g$ |
| 8 | $250+g+250-g=500$ | $500+h$ |
| 9 | $500+h+500-h=1000$ |  |

Thus, on the ninth move, the first player wins.

Answer: the first.