Bump......

I'll show that the person that plays first has the winning strategy, which is as follows: He always chooses the smallest $3$ remaining numbers, different from $1$ , because $1$ can never be picked. Now I'll show that he can always choose them, i.e. they will always form the sides of a triangle. Assume that $k$ moves have passed and it's the first persons turn($k<67$), i.e. he has removed out $3k$ numbers and the other dude has removed $3k$ numbers as well. Note that there are at least $4$ numbers left. Let the $2$ smallest numbers remaining be $a$ and $b$, $a<b$. Assume that the second person has removed $t$ numbers smaller than $b$ and so $3k-t$ numbers larger than $b$. The first person has removed $3k$ numbers smaller than $a$, so $a\geq3k+2$ and $b=3+3k+t$. Now the first person can't choose a 3rd number that forms a tringle with $a$ and $b$ iff all the numbers from $b+1$ to $b+a-1$ are removed. Because $b+3k-t=3+6k<400$, then there exists a number larger than $b$ that isn't removed and also the second person has removed $3k-t$ numbers larger than $b$, but $a-1\geq 3k+1 >3k-t$, so the first person can indeed choose another number that forms a triangle, so we choose the smallest such number. Now that we have shown his strategy we just have to note that $400-1=399=6.66+3$, so the first person can guarantee himself playing $67$ moves, leaving the second person with the number $1$, unable to move, thus winning.

We claim that the starting player wins. He does it with always choosing the $3$ smallest numbers except $1$. Lets suppose after some moves for $k,l\geq 1$, smallest $3$ numbers are $x,x+k,2x+k+l-1$ (except $1$). Then, the second player has removed at least $k + 1 + x + l$ numbers. So first player also removed at least $x + k + l + 1$ numbers as well. But as the smallest number is $x$ and first player always removes the smallest $3$ numbers, first player must have removed at most $x-2$ numbers. But as $x-2 < x + k + l + 1$, this is a contradiction. So first player can always choose the $3$ smallest numbers which form a triangle. At the end, after first player moves, there will be $1$ number left (which is $1$) and so the second player loses, first player wins.