This problem was proposed by Federico Bach and myself, Santiago Rodriguez, both of us from Colombia. I hope that you enjoy our problem.

@above thanks, I added you guys. Its a great problem

We may substitute any given candy of type $k$ with a candy of type $k-3$ while leaving the rest of the candies unchanged, through the following sequence of operations: $$(k)\to(k-1,k+1)\to(k-2,k,k+1)\to(k-3,k-1,k-1,k+1,k+1)\to(k-3).$$By trivial induction for which the previous result is the base case, we may substitute any given candy in the row of type $k$ by a candy of type $k-3t$, where $t$ is any positive integer, and we use cyclic notation modulo $n$. Now, if $n$ is coprime with $3$, an integer $t$ exists such that $k-3t$ takes any desired remainder modulo $n$, or given two consecutive candies $k,k'$, a $t$ exists such that $k-3t\equiv k'\pmod n$, and we may perform operations leading from $(k,k')$ to $(k-3t,k')$, and then eliminate both through an operation of type 2. If $m$ is initially odd, we perform first an operation of type 1, resulting in an even $m$. Once $m$ is even, $\frac{m}{2}$ sequences of transformations $(k,k')$ to $(k-3t,k')$ followed by elimination of the two consecutive candies of the same type results in all candies being removed from the table. Assume now that $n$ is a multiple of $3$, and let $m=m_1+m_2+m_3$, where $m_i$ is the number of candies of any type $k$ such that $k\equiv i\pmod3$. Note that since $n$ is a multiple of $3$, regardless of the value of $k$, when an operation of type 1 is perfomed on a candy of type $k$, one of $m_1,m_2,m_3$ decreases by $1$, and the other two increase by $1$. On the other hand, an operation of type 2 results in one of $m_1,m_2,m_3$ decreasing by $2$, and the other two remaining unchanged. Therefore, in each operation, either all of $m_1,m_2,m_3$ switch their parities, or all their parities remain unchanged. As a result, if initially not all of $m_1,m_2,m_3$ have the same parity, no sequence of operations may make them all simultaneously even, and thus not all three of them may be made simultaneously zero. Or, some candy will always remain on the table. We conclude that it is possible to clear the table for any initial configuration of candies iff $n$ is coprime with $3$.

BTW, beautiful problem @Supertinito!!!

I quite like this problem We claim the answer is all $n$ not divisible by $3$ AFTOC $3$ divides $n$. Let each type of candy be its residue mod $n$. Notice these residues can be partitioned mod $3$ as well. Now, if there is a candy of residue $a$ and we operate on it, a piece of candy is added of residue $a+1$ and $a-1$ and removed from $a$. Thus, the number of candy of each residue mod $3$ switches parity. So, if we start with one candy of residue $1$ mod $n$ and no other candies, the number of candies on each residue mod 3 will be odd, even, even, or even, odd, odd, and never even, even, even. Proving that $0$ candies is impossible. Now, we assume $3$ does not divide $n$. Consider the following algorithm to make the number of candies on each residue even (at which point completion is trivial). We will now indicate residues with an even number of candies as $0$ and odd as $1$. First, we arbitrarily operating on $1$s with a neighbor which is also $1$. This strictly reduces the number of $1$s so we will eventually get no neighboring $1$s. Now, since $3$ does not divide $n$ we can choose two $1$s with no $1$s between them where the number of zeros between them plus 1 is not divisible by $3$. Now, considering the following algorithm \[00010 \rightarrow 00101 \rightarrow 01011 \rightarrow 10111 \rightarrow 10000 \]gets one of the two $1$s closer by $3$. Thus, by repeating this algorithm we will get our two $1$'s next to each other thereby decreasing the number of $1$s when operating on either one of them. Now, we have one $1$. Since this proof is long enough as is, I will not prove it, but the construction is not hard to find. It involves filling almost every residue with $1$s and partitioning them into groups of $3$ to kill all of them. Thus we are done. $\blacksquare$

Cute. The answer is $3 \nmid n$. If $n$ is not divisible by $3$, notice that from a single candy of type $1$ Celeste can perform the following moves: \[ 1 \mapsto 02 \mapsto 013 \mapsto 00224 \mapsto 4. \]In other words, Celeste can apply $+3$ shifts to any candy type. So if $3 \nmid n$, this is the same as being able to transform the type of any candy on the row however she wishes without affecting the rest of the row. Clearly that means the task is possible. Conversely, suppose $3 \mid n$. Define the Klein four group \[ G = \left\{ 1, a, b, c \right\} \cong {\mathbb Z}/2 \oplus {\mathbb Z}/2 \]with multiplication table $a^2=b^2=c^2=1$ and $bc=cb=a$, $ca=ac=b$, $ab=ba=c$. We'll assign candies of types $0$, $1$, $2$ mod $3$ a label $a$, $b$, $c$. Then any row can be interpreted as a string of elements of $G$, and thus (by group multiplication) an element of $G$. By construction, the operation leaves this element invariant. This means Celeste cannot complete the task starting from a single candy, because the end state is the empty string which has product $1$.

The answer is all $3 \nmid n$. Proof of Necessarity: Suppose that $3 \mid n$. Let $a$ be the number of cards that have $0$ mod $3$ labels, and define $b, c$ similarly. Observe that originally, $a, b, c$ are of the same parity. Furthermore, every move either does not change the parity of any of the three at all or changes all of them. However, when all cards have been removed, $(a, b, c) = (0, 0, 0)$. Considering a configuration when there is only a single card with label $1$, this is a clear contradiction. Proof of Sufficiency: We induct from $n-3$ to $n$. The base cases $n=1$ and $n=2$ can be checked manually. For the inductive step, it suffices to show that we can decrease the index of all cards indexed $k > 3$ to $k-3$. To do so, use the series of operations $$(k) \to (k-1, k+1) \to (k-2, k, k+1) \to (k-2, k-1, k+1, k+1) \to (k-2, k-1) \to (k-3, k-1, k-1) \to (k-3).$$Thus, we can reduce all cases with maximum label $n+3$ to those with maximum label $n$, so the induction is complete.

We call the first type of operation a split and the second type a take. We claim that the answer is all numbers $n$ such that $\boxed{n \not\equiv 0 \pmod{3}}$. We have $3$ cases to consider. $\textbf{Case 1:} n \equiv 0 \pmod{3}$ Consider the caser of $m=1$. We claim that for any such case the result is impossible. Consider the types $\pmod{3}$ in which we have $3$ groups. Consider the parities of the number of total number of candies in each group. They must obviously all be even to achieve what we want. However is we split one candy, then each group is changed to the opposite parity, and if we take a pair of candies the parity of every group remains the same. Thus we can see the result. $\textbf{Case 2:} n \equiv 1 \pmod{3}$ Let there be $3k+1$ types of marbles. We claim that we can make any number decrease by $1$ without effecting the other numbers. We start with a number $x$ and repetitively split the lower number of the other split (the first split is of $x$). We do this $3k$ times. Then we have $3k$ consecutive single marbles and the $x-1$ marble. For a single group of $3$ we can split the middle one and take the two pairs that it makes. We can make a diagram of moves with each new line representing a position after a certain move. For example, when we have $x = 5$ and $n = 7$ we get: $5$ $4, 6$ $4, 5, 0$ $4, 5, 6, 1$ $4, 5, 6, 0, 2$ $4, 5, 6, 0, 1, 3$ $4, 5, 6, 0, 1, 2, 4$ $4, 4, 6, 6, 0, 1, 2, 4$ $6, 6, 0, 1, 2, 4$ $0, 1, 2, 4$ $0, 0, 2, 2, 4$ $2, 2, 4$ $4$ Now we try and prove that from this we can delete the whole sequence. If the number of terms is even we can simply subtract $1$ to the first one until it equals the second for every pair of $2$ numbers. If the number of terms is odd. We can simply delete the all of the terms except $1$ in the way above, and then split the last one and delete that pair. $\textbf{Case 3:} n \equiv 2 \pmod{3}$ This is quite similar to the above case. Let there be $3k+2$ types of marbles. We claim that we can make any number increase by $1$ without effecting the other numbers. We start with a number $x$ and repetitively split the lower number of the other split (the first split is of $x$). We do this $3k+3$ times. Then we have $3k+3$ consecutive single marbles and the $x+1$ marble. For a single group of $3$ we can split the middle one and take the two pairs that it makes. We can make a diagram of moves with each new line representing a position after a certain move. For example, when we have $x = 5$ and $n = 8$ we get: $5$ $4, 6$ $4, 5, 7$ $4, 5, 6, 0$ $4, 5, 6, 7, 1$ $4, 5, 6, 7, 0, 2$ $4, 5, 6, 7, 0, 1, 3$ $4, 5, 6, 7, 0, 1, 2, 4$ $4, 5, 6, 7, 0, 1, 2, 3, 5$ $4, 5, 6, 7, 0, 1, 2, 3, 4, 6$ $4, 4, 6, 6, 7, 0, 1, 2, 3, 4, 6$ $6, 6, 7, 0, 1, 2, 3, 4, 6$ $7, 0, 1, 2, 3, 4, 6$ $7, 7, 1, 1, 2, 3, 4, 6$ $1, 1, 2, 3, 4, 6$ $2, 3, 4, 6$ $2, 2, 4, 4, 6$ $4, 4, 6$ $6$ Now we try and prove that from this we can delete the whole sequence. If the number of terms is even we can simply add $1$ to the first one until it equal the second. If the number of terms is odd. We can simply delete the all of the terms except $1$ in the way above, and then split the last one and delete that pair. Combining these cases we see the result. $\blacksquare$

very nice problem We claim that the answer is all positive integers except for multiples of $3$. When $n \equiv 0 \pmod{3}$, fix $m = 1$. Note that we can only delete cards when they are equivalent modulo $3$, which means the parities of the number of cards in each respective modulo $3$ must all be even. But when we perform a replacement, each of the parities are changed by $1$. Thus, if we start so that one of the parities is odd, there will always be an odd parity, which means no cards will ever be deleted. When $n\not\equiv 0\pmod{3}$, we can perform the transformation, \begin{align*} k&\rightarrow k-1,k+1 \\ &\rightarrow k-2,k,k,k+2 \\ &\rightarrow k-2, k+2 \\ &\rightarrow k-2, k+1, k+3 \\ &\rightarrow k-2, k, k+2, k+3 \\ &\rightarrow k-2, k-1, k+1, k+2, k+3 \\ &\rightarrow k+3 \end{align*}Similarly, we can also perform the transformation to get $k \rightarrow k-3$. Note that, we have \begin{align*} k &\rightarrow k-1, k+1 \\ &\rightarrow k-2, k+2 \end{align*}So if $n\equiv 1\pmod{3}$, we can get \begin{align*}k &\rightarrow k-3(\frac{n-1}{3})-1, k+3(\frac{n-1}{3})+1\\ &= k-n, k+n \\ &= k, k\end{align*}or the other way around using $(k-2, k+2)$. This means that we can individual delete a candy if $n\not \equiv 0\pmod{3}$, so we can simply perform this algorithm for all candies to have the table be empty, which thus concludes the proof.

remarks

The answer is all n with $3\nmid n$. For convenience, (1) means the first operation and (2) means duh. Notice if 3|n take m=1 for simplicity, and make three groups of candies of types taken mod 3. It starts out with odd,even,even for parities, which we'll denote o and e as the # of odd and even groups. We want to end with even,even,even (all 0s), which is o=0 and e=3. But notice (1) changes all parities, which keeps o>0 and e>0 if they were already >0 (since (o,e)=(1,2)<->(2,1)), while (2) doesn't change o or e. Hence o can never reach zero by these operations. Now if n is not divisible by 3, we can proceed by induction. n=1 is immediate, n=2 you can take any type 2 and exchange for type 1 and type 3=type 1 to cancel everything in type 1s. Now, the algorithm is $$k->k-1,k+1\rightarrow k-2,k,k,k+2\rightarrow k-2,k+2\rightarrow k-3,k-1,k+2\rightarrow k-3,k-2,k,k+2\rightarrow k-3,k-2,k-1,k+1,k+2\rightarrow k-3,k-2,k-2,k,k+1,k+2\rightarrow k-3,k,k,k+2,k+2\rightarrow k-3$$reduces maximum type n to maximum type n-3, hence we are done. $\blacksquare$ honestly a nice problem, had invariants, algorithm but pretty easy

really good problem We claim the answer is all $n$ such that $n\equiv 1,2\pmod{3}$. Assume FTSOC $n\equiv 0\pmod{3}$. When considering the number of terms, $a,b,c$ for each respective modulo $3$, we can say that they must all be even. This is because in order to be left empty, and the fact that $n\equiv 0\pmod{3}$, we must take two out each time, implying that the parity of $a,b,c$ is all even. However, when starting out with and odd,even,even parity, if we replace it, we always change the parity of each by $1$, which never gets an all even parity. We now prove that everything else works. Consider the following transformation: \[k\rightarrow k-1,k+1\rightarrow k-2,k,k+1\rightarrow k-2,k-1,k+1,k+1\rightarrow k-2,k-1\rightarrow k-3,k-1,k-1\rightarrow k-3.\]This implies that it suffices to look at $k\pmod{3}$. When we subtract any multiple of $3$ from it, it will become a different modulo $3$, implying that we can pair terms out leaving the table empty, as desired.

We claim that the answer is only $3\nmid n$ work. First we prove that all $3\nmid n$ work. We can go from some $a$ to $a+3$ like this: \[a\rightarrow a-1, a+1 \rightarrow a-1, a, a+2 \rightarrow a-1, a-1, a+1, a+2\rightarrow a+1, a+2\rightarrow a+1, a+1, a+3\rightarrow a+3\]WLOG, the sequence currently an even number of candies (because otherwise, we can get an even number of candies by applying the first process). Now, that means for any candy $a_i$ in the sequence of candies $a_1, a_2, \ldots, a_k$, since $n\not\equiv 0\pmod 3$, we have a solution to $a_i + 3x \equiv a_1\pmod n$ for all $a_i$, so we can turn all the candies into $a_1$, and then eat all of them. Now we show a counterexample for $3\mid n$. Let $M_i$ be the number of candy types that are $i\mod 3$ for $i=0, 1, 2$. Notice that each time we use the first option with $a\equiv i\pmod 3$, then $M_i$ is now $M_i-1$ with $M_{i-1}$ and $M_{i+1}$ are now $M_{i-1} + 1$ and $M_{i+1} + 1$, respectively. In the first case, the parity of all $M_i$ stay the same, and in the second case, the parity of all $M_i$ change. So, if we start with $M_0 = 1, M_1 = 0, M_2 = 0$, then we will never be able to have all $M_i$ be even, meaning we cant ever get to $0$ candies. really liked this one :')))

The answer is $\boxed{3 \nmid n}$. First, we will show that for all $n$ such that $3 \nmid n$, Celeste can complete the task. Claim 1: Celeste can turn a candy of type $k$ into a candy of type $k+3$ though a series of operations Proof: WLOG we start with one candy of type $1$. The following algorithm turns it into a candy of type $4$. \[1 \to 02 \to 013 \to 0023 \to 00224 \to 4. \ \square\] It is clear that the task is possible for $3 \nmid n$. Now, we will prove that the task is not always possible to complete for $3 \mid n$. We claim that if $m=1$, it is impossible. Claim 2: Let $x,y,z$ be the number of candies with type $0,1,2 \pmod{3}$ respectively. The parities of $x,y,z$ either all toggle or none of them toggle Proof: This can be manually checked by looking at the operations: The first one toggles all of them and the second one toggles none. $\square$ Starting with only one candy on the board, it is impossible to eventually have none left by Claim 2, finishing the proof.

The question relies on the following claim: Claim: It is possible to transform a candy of type $k$ to a candy of type $k+3$. \[(k) \rightarrow (k-1, k+1) \rightarrow (k-1, k, k+2) \rightarrow (k-1, k-1, k+1, k+1, k+3) \rightarrow (k+3). \quad {\color{blue} \Box}\] Therefore if $n$ is not a multiple of 3, we have $\gcd(n, 3)=1$, so we can transform all of the candies into a single type, say type 1. If we have an odd number of candies, replace one of the candies with a candy of type 0 and a candy of type 2, both of which can transformed back to a candy of type 1, leaving us with an even number of candies with type 1. At this point, we simply eat all of the candies in pairs. If $n$ is a multiple of 3, this algorithm no longer works, so we claim it is impossible for at least one configuration. Denote $a$, $b$, and $c$ as the parities of the number of candies with type 0, 1, and 2 modulo 3, respectively. Each operation will either completely flip or completely preserve the triple $(a,b,c)$. Hence if our configuration begins with $(E,E,O)$, it is impossible to get to $(E,E,E)$, which is the desired ending. Thus our answer is $\boxed{n \equiv 1,2 \pmod 3}$. $\blacksquare$

Consider the following. $$3 \rightarrow 24 \rightarrow 134 \rightarrow 0234 \rightarrow 02244 \rightarrow 0.$$Thus, we can change all candies to all other types that have the same residue $\pmod 3.$ Note that when $n \equiv 1,2 \pmod 3,$ we can change all candies to the same type as $\gcd(1,3)=\gcd(2,3)=1.$ If we have an even number of candies, we're done, if there are an odd number, we eat all but $1,$ split it into $2$ candies, so now we're done. \newline We now prove that if $n|3$ then it's not always possible. Note that eating or splitting does not change the residue of the sum of all types present, with each type being added the number of times it appears. Suppose there is a type $2$ and a type $3,$ the residue of the sum is thus $2,$ which $\neq 0,$ the sum of $0$ candies, and we're done.

The answer is all $n$ so that $3$ does not divide $n$.. We claim that we can turn any candy of type $k$ into a candy of type $k+3$. $k \to k-1, k+1 \to k-1, k, k+2 \to k-1, k, k+1, k+3 \to k-1, k-1, k+1, k+1, k+3 \to k+3$. And then if $(3, n) = 1$, we can continuously repeat this process to turn any $k$ into $k + 3d = m$ for arbitrarily $m$, since $3$ is relatively prime to $n$. From here, it is easy to delete all candies as we can split $k$ into $k-1$ and $k+1$, and turn $k-1$ to $k+1$ and delete the two. If $3 \mid n$, we divide the candies modulo $3$ by type. Then each move either swaps the pairities of the numbers of each candy in each class(modulo $3$) or keeps them all the same(deleting $2$ candies). In an ending state, the pairities of the number of each candy in each modulo $3$ class are all the same, however if we start with a configuration where the pairities are different then we will never reach the ending state.

The answer is that Celeste can leave the table empty for any value of $m$ and any configuration of candies on the table if and only if $gcd(n,3)=1$. First, we show that is indeed possible to leave the table empty when $3 \nmid n$. Notice that if it is possible to leave the table empty starting with $m=1$ candy then, Celeste simply has to do the required steps on each individual candy in a collection of $m>1$ candies and he will be done. Further, we only need to consider this candy to be of Type 1 since the steps for other starting types will be entirely the same but shifted $\pmod{n}$. Algorithm : It is possible to convert a candy of Type $k$ to Type $k+1$ for $n \equiv 2 \pmod{3}$ and convert a candy of Type $k$ to Type $k-1$ for $n \equiv 1 \pmod{3}$ via the following algorithm. Simply consider the following sequence of steps. \begin{align*} &k \\ k-1 \ & \ k+1\\ k-1 \ & \ k \ k+2 \\ k-1\ k-1 & \ k+1 \ k+1 \ k+3 \\ &k+3 \end{align*} Proof : This sequence of steps can convert a candy of Type $k$ to Type $k+1$. Consider $n=3i+1$, note that we can convert a candy of Type $1$ to $3i+1=n$ as needed. Consider $n=3i+2$, note that we can convert a candy of Type $1$ to $3i+1$ and then to Type $2 = 1+1$ as needed. Thus, we have our claim. Now, note that one can convert a candy of Type $1$ to a candy of Type $n$ and Type $2$. And then, using the above algorithm, in both cases we can convert the candy of Type $2$ into Type $n$. Thus, we will be done. Now, we show that it is indeed impossible to eat all candies for $n\equiv 0 \pmod{3}$. Let $T_i$ be the number of candies of Type $i$. Further, let $S_i$ ($i=1,2$) be the total number of candies of Type $j$ such that $j \equiv i \pmod{3}$. Now, notice that in the beginning $S_1 \equiv 1 \pmod{2}$ and $S_2 \equiv 0 \pmod{2}$ for the starting configuration of a single candy of Type 1. Consider $1<i<n$ with $n\equiv 1 \pmod{3}$. Then, notice that doing a move on this a candy of Type $n$ will have the effect of changing the sums into $S_1-1$ and $S_2+1$ flipping the parity of both thus, after the move atleast one of the two will always be odd (since it was in the start). Thus, one of $T_1,\cdots$ must be odd. Similarly, when $n \equiv 2 \pmod{3}$ with $S_1+1$ and $S_2-1$. If $n \pmod{3}$, then the change is into $S_1+1$ and $S_2+1$, thus, again the parity of both changes, ensuring atleast of one them remains odd. One can check that if $n=3k$, then a move done on a candy of Type $1$ or $n$ will also have the same effect. Thus, we have on invariant and atleast one of $T_1, \cdots , T_n$ will be odd, which means that it is impossible to delete all the candies (since deletion deletes two of each type making no change to the parity). So, indeed the answer is as claimed.

Answer is all integer $n$ such that $3$ dose not divide $n$. Claim: For $n= 3k+1$ or $3k+2$ we can remove each candy individually \paragraph {Construction} We show that we can delete each candy. So consider we only have candy of type $k$. For terminology if we say $k-p$ that suppose to mean candy of type $k-p$ mod $n$ Lemma: From two candy of type $(k-a,k+a)$ we can go to $(k-(a-3),k+a-3)$ We apply operation (1) and (2) repeatedly on selective candys \begin{align*}(k-a,k+a) &\stackrel{(1)}{\rightarrow} (k-a-1,k-a+1),(k+a-1,k+a+1) \\ &\stackrel{(1)}{\rightarrow} k-a-1,(k-a,k-a+2),(k+a-2,k+a),k+a+1 \\ &\stackrel{(1)}{\rightarrow} k-a-1,(k-a-1,k-a+1),k-a+2,k+a-2,(k+a-1,k+a+1),k+a+1 \\ &\stackrel{(2)}{\rightarrow} k-a+1,k-a+2,k+a-2,k+a-1 \\ &\stackrel{(1)}{\rightarrow} k-a+1,(k-a+1,k-a+3),(k+a-3,k+a-1),k+a-1 \\ &\stackrel{(2)}{\rightarrow} k-a+3,k+a-3 \end{align*} Note that from $(k-3,k+3)$ we can go to non (so we eat both candy and also other candy made by this two candy). This is because from $(k-3,k+3)$ we go to $(k,k)$ by lemma and then we eat up both candy. Hence observe in proof of lemma that from \[( k-a+1,k-a+2,k+a-2,k+a-1) \stackrel{(1)}{\rightarrow} (k-a+3,k+a-3)\](as we use this repeatedly later). Observe $k-n \equiv k+n ($mod $n)$. On single candy $k$ if we apply only operation (1) on rightmost and leftmost candy type we get \begin{align*} k &\stackrel{(1)}{\rightarrow} k-1, k+1 \\ &\stackrel{(1)}{\rightarrow} k-2,k,k,k+2 \\ &\stackrel{(1)}{\rightarrow} k-3,k-1,k,k,k+1,k+3 \\ &\cdots \cdots \\ &\stackrel{(1)}{\rightarrow} k-n,k-n+2, \cdots k-2,k-1,k,k,k+1,k+2, \cdots k+n-2,k+n \end{align*}So this sequence have all number from $k-n+2$ to $k+n-2$ and at end we have $k-n$ and $k+n$. Now we remove $k,k$, then we convert $k-2,k-1,k+1,k+2$ into $k,k$ and remove it again. So each time we get $$k-3r-2,k-3r-1,k-3r,k+3r,k+3k+1,k+3r+2$$we can convert $k-3r,k+3k$ into $(k,k)$ by applying lemma as many times, then remove it. And remaining $k-3r-2,k-3r-1,k+3k+1,k+3r+2$ can convert into $k-3r,k+3r$ which we remove again. So as per $k-n+2$ is of type $k-3r$ or $k-3r-2$ we can remove all block of $k-n+2$ to $k+n-2$ by above procedure. For that we should have $ n \equiv 1,2 ($mod $3)$. and after remove everything as $k-n = k+n$ in mod $n$ we can remove both of them. Similarly doing this for all candy , we can remove all $m$ candies. Contradiction for case 3 divides n If $n=3k$, Consider $E$ as sum of number of candy of type $3,6, \cdots 3k-3, 3k$ (all candy of type $0 \equiv $mod $3$). If $A$ denote number of candy present on row at every moment. define $S= A- E$ Main observation is when we remove any candy of type $0 \equiv $ mod $3$ by operation (1) , we increase $S$ by $2$ and if we remove other type of candy by operation (1) then $S$ does not change. For operation (2) , we only can decrease $S$ by $2$. Hence $S \equiv ($mod $2)$ never change. If we take only $1$ candy of type $1$, then $S = 1$ and hence it will never become $0$. Therefore we never can have $0$ candy on row for $3|n$.

$\boxed{3 \not | n}$ Consider the following operations: $$A_n \implies A_{n-1} A_{n+1} \implies A_{n-1} A_n A_{n+2} \implies A_{n-1} A_{n-1}A_{n+1} A_{n+2} \implies A_{n+1} A_{n+1} A_{n+3} \implies A_{n+3}$$Therefore if $n$ is not divisible by $3$ then we can easily delete all numbers because we can transform any candy of one type into another. If $n$ is divisble by $3$, then the process is impossible. The key is to split the number of candies into the three $\pmod{3}$ residue classes. The number of candies in each of these classes when taken $\pmod{2}$ are constant or replaced by its complement. This is trivial to see by the conditions in the problem. But if we start with $(1, 0, 0)$ in some order then we cannot reach $(0, 0, 0)$, proving impossibility.