A Toothpick Game between Calvin and Brian!

Calvin has a winning strategy exactly when $n\not\equiv0,2\pmod{7}$ . It's easy to check that this holds for the first few values of $n$ . Inductive, assume this is true for all $n<k$ . Take $n=k$ . If $k\neq0,2\pmod{7}$ , then for some $m\in\{1,3,4\}$ , $k-m\equiv0$ or $2\pmod{7}$ . By inductive hypothesis, Brian loses. However, if $k\equiv0$ or $2\pmod{7}$ , then for any choice of $m$ , $k-m\not\equiv0,2\pmod{7}$ . Therefore, Brian wins.

Thus, Calvin has a winning strategy when $n=31,32,33,34$ but not $35$ , so the answer is $\boxed{130}$ .

Unfortunately, $100\equiv 2\pmod{7}$ , so Calvin does not have a winning strategy when there are $100$ toothpicks.

Similarly, if they are allowed to remove $1,2,4$ or $1,3,6$ toothpicks, Calvin has a winning strategy for $n\not\equiv0\pmod{3}$ , $n\not\equiv 0,2,4\pmod{9}$ , respectively. In either case, Calvin has a winning strategy when there are $100$ toothpicks.

We can generate two lists of numbers iteratively, one where Calvin can have a winning strategy and one where he can't. Where list $W$ is the winning numbers and $L$ is the losing numbers, we start our lists as the following:

$W=\{1,3,4\}$

$L=\{2\}$

Clearly Calvin can win on anything in the win group simply by subtracting $1$ , $3$ or $4$ . He can't win with $2$ because he can only take away $1$ and then Brian would take the last toothpick.

Our strategy goes as follows. If Calvin can subtract $1$ , $3$ or $4$ toothpicks to get a number on the losing list, then that would be a winning strategy for Calvin and that number is added to the winning list. If he can't, then that number is added to the losing list. Using this algorithm we can generate the lists as follows:

$W=\{1,3,4,5,6,8,10,11,12,13,15,17,18,19,20,22,24,25,26,27,29,31,32,33,34\}$

$L=\{2,7,9,14,16,21,23,28,30,35\}$

It's clear that a pattern emerges, showing that all our numbers on the losing list are of the form $7k$ and $7k+2$ . For $n\in[31,35]$ , there are $4$ winning strategies: $31$ , $32$ , $33$ & $34$ , meaning that our answer must be:

$31+32+33+34=\boxed{130}$