## Task 2.

Three friends found $N$ coins with a total mass of 300 grams. The mass of each coin is a natural number. They decided to divide the coins into three piles of equal mass. For which $N$ is this definitely possible, regardless of the mass of individual coins?
## First Solution.

The smallest such $N$ is 201.

It is possible that the masses of all but one of the coins are 1 gram, and the remaining coin has a mass of $301-N$ grams. If $N \leqslant 200$, then the mass of the heaviest coin would be at least 101 grams, making it impossible to distribute the coins into three piles of 100 grams each.

Let's prove that for $N=201$ coins, it is possible to distribute the coins into three piles of equal mass, regardless of the masses of the coins. Let $a_{1} \geqslant a_{2} \geqslant a_{3} \geqslant \cdots \geqslant a_{201}$ be the masses of these coins in grams.

First, we conclude that at least 101 of these coins have a mass of one gram. Otherwise, if the mass of $t \leqslant 100$ coins is one gram, and the mass of $201-t$ coins is at least 2 grams, the total mass of all coins would be at least $t \cdot 1 + (201-t) \cdot 2 = 402 - t \geqslant 302$ grams, leading to a contradiction. Therefore, we have

$$
a_{101} = a_{102} = \cdots = a_{201} = 1
$$

Place 100 coins of mass one gram on the first pile.

For the other two piles, consider the sums

$$
a_{1}, a_{1} + a_{2}, a_{1} + a_{2} + a_{3}, \ldots, a_{1} + a_{2} + \cdots + a_{100}
$$

All 100 sums have distinct values, ranging from at least 1 to at most 199 (since the sum of the remaining masses is 101, as each coin has a mass of 1). If one of these sums is equal to 100, place the coins with those masses on the second pile, and the rest on the third pile, completing the construction. If none of these sums is equal to 100, in particular, none of the sums is divisible by 100, then by the Pigeonhole Principle, there exist two sums that give the same remainder when divided by 100. Consider the sum obtained by subtracting these two expressions. It is divisible by 100, and it is greater than zero and less than 200, so it must be 100. Place the coins with the masses that are the addends in this sum on the second pile, and the rest on the third pile. This completes the construction in this case as well, thus proving the claim.