Really nice problem!, initially I thought this would be very easy, because it seemed obvious that the sages could be able to avoid just one number as the eventual sum. But as it turns out, I think that if the condition was changed to the sum of the chosen numbers cannot be $200$ or $201$, then I think the guard can actually ensure that the sages cannot guarantee a win. Also, I've probably written the solution badly because I didnt know how better to explain it All sages except for the last use the following strategy: 1. If the previous two numbers chosen were $1, 1$, then they choose the smallest number available 2. If the previous two numbers chosen were $2, 1$ or $3, 1$, then they choose the number $1$ if possible, else choose $3$ 3. In all other cases, choose the smallest number available The last sage has the following strategy: 1. If the previous number chosen was $1$, then choose the smallest number available 2. Otherwise, choose the number $1$ if possible, else choose the number $3$ Now, I'll show that this strategy works. Define the deviation after $k$ sages have passed as $2k - S_k$ where $S_k$ is the sum of the $k$ numbers chosen so far. So, choosing $2$ keeps deviation same, choosing $3$ reduces it by $1$ and choosing $1$ increases it by $1$. The sages only lose if the deviation at the end is $0$ Observe that the deviation at any point is always nonnegative. This is because the only thing that can reduce it, a $3$, is only chosen if a $1$ was chosen the previous move and so the $31$ balance each other out. Also, see that the guard cannot let $1$ be chosen twice in a row since that would cause the deviation to become at least $2$ and from that point on, the deviation does not reduce any further until $99$ sages are done choosing. The last sage can reduce the deviation by at most $1$ and so the deviation at the end is at least $1$ and so the sages win. So now, observe that if $1$ is ever chosen, then the next sage, according to the strategy will pick $3$ since the guard has to not let him pick $1$, according to the previous paragraph. So, the net deviation is $0$. So, we can see that after $98$ or $99$ sages are done, the net deviation is still $0$. First, suppose net deviation after $98$ sages is $0$ but not after $99$. So, the 99th sage picks the number $1$, making the deviation $1$. But now, according to the strategy, the last sage picks the smallest number, which is $1$ or $2$, which cannot decrease the deviation and so the deviation at the end is $\ge 1$ and the sages win Finally, the last case is when after $99$ sages, the deviation is $0$. Obviously, this means that the 99th sage did not pick $1$ since otherwise this means that the deviation after $98$ sages was $0$, which is the previous case. So, according to the strategy, the last sage picks $1$ or $3$, which makes the deviation nonzero and so the sages win.

Let $s_1,s_2,\ldots,s_{100}$ be the sages ; $T(s_i)$ be the set given by the guard to $s_i$ ; $N(s_i)$ number chosen by $s_i$ ; $B(s_i)$ be $N(s_{i-2})N(s_{i-1})$. By convention, we define $N(s_{-2}),N(s_{-1}) = 2$ (so that $B(s_1),B(s_2)$ are well defined). The strategy for $s_i$ is following: If $1 \le i \le 99$, then If $T(s_i) = \{1,3\}$, take $N(s_i) = 1$. $T(s_i) = \{2,a\}$ with $a \in \{1,3\}$. If $B(s_i)$ is $21$ or $31$, take $N(s_i) = a$. Otherwise, take $N(s_i) = 2$. If $i=100$, then If $T(s_{100}) = \{1,3\}$, take $N(s_i) = 1$. $T(s_{100}) = \{2,a\}$ with $a \in \{1,3\}$. If $B(s_i)$ is $21$ or $31$, take $N(s_i) = 2$. Otherwise, take $N(s_i) = a$. We show this strategy works now. Firstly, in the sequence $\mathcal S = N(s_1),N(s_2),\ldots,N(s_{99})$, change all instances of $1,3$ by $2,2$. We can do this as the sum remains same while our strategy for $s_i$ (with $1 \le i \le 99$) was same for $B(s_i)$ to be $\{13,22\}$ and $\{31,21\}$ (so our strategy isn't disturbed). By strategy, it isn't hard to see in $\mathcal S$ cannot contain a $2,1,2$. Also, as before each $3$ we had a $1$, so in the sequence there are no more $3$'s in $\mathcal S$ now. So $\mathcal S$ now looks something like $$ 2,\ldots, 2, \underbrace{1,\ldots,1}_{\text{at least two}}, 2 , \ldots , 2, \underbrace{1 , \ldots , 1}_{\text{at least two}}, 2 , \ldots, 2, \cdots $$Suppose $k$ is the number of $1$'s among $s_1,\ldots,s_{99}$. We have several cases: $k = 0$. Then $\mathcal S$ must be $2,2,\ldots,2$. Then $s_{100} \ne 2$ by strategy, so sum isn't $200$. $k=1$. Then $\mathcal S$ must be $2,2,\ldots,2,1$. Then $s_{100} = 2$ by strategy, so sum isn't $200$. $k \ge 2$. Then even if $s_{100} = 3$, the sum still is still $< 200$. This completes the proof. $\blacksquare$

An interesting and surprisingly difficult problem. Shift everything by $-2$, so the set is $\{-1,0,+1\}$ and the sages wish to avoid $0$. The first $98$ sages employ the following strategy: if the last number chosen was $1$, but the number chosen before that was not $1$, then choose the number $3$ if possible, otherwise choose the number $1$. Otherwise, choose the smallest number available, which is at most $0$. If a $+1$ is ever chosen in the first $99$ numbers, the number immediately before it must be a $-1$. On the other hand, every $-1$ not adjacent to a $+1$ must be adjacent to a $-1$, except for possibly the $99$-th number chosen. Therefore, unless the $99$-th number is a $-1$, the sum of the first $99$ numbers is either $0$ or at most $-2$, otherwise it is at most $-1$. The $100$-th sage then looks at the $99$-th number chosen. If it is a $-1$, then they choose the smallest number, and the final sum is negative. Otherwise, they choose an odd number, which is always possible, making the final sum either at most $-1$ or $1$. In either case, they win. $\blacksquare$