I have the same answer as ioanandrei. The small cases ($n=3,4$) are left to the reader's pleasure. At a certain moment, the vector $(a_1,...,a_n)$ shows that the kid $i$ has $a_i$ candies. We split the problem in two cases, $n$ odd and $n$ even. If $n$ is odd, the main idea is that if for an $i$ we have $a_i\ge 1$, in two steps we can obtain $a_i\rightarrow a_i-1,\ a_{i+2}\rightarrow a_{i+2}+1$ and every other $a_j$ remains unaffected. Suppose wlog $i=1$ for the simplicity of the explanation. We pair $a_k$ with $a_{k+1}$ for all $2\le k\le \dfrac{n-1}{2}$ and for the two steps we perform the following: 1) if both $a_k$ and $a_{k+1}$ are at least one, take $1$ candy from $a_k$ and give it to $a_{k+1}$ and viceversa. Doing this two times obviously leaves $a_k$ and $a_{k+1}$ unchanged. 2) if one of them is zero, say $a_k=0$, then take all the candies from child $k$ and give them to child $k+1$. This way, child $k$ has now $a_{k+1}$ candies and child $k+1$ has $a_k$ candies. Doing this one more time leaves everything as it was before. In the first step take $a_2$ and give it all to child $1$ and take all of $a_1$ and $a_3$ and give them to child $2$, so $(a_1,a_2,a_3)\rightarrow (a_2,a_1+a_3,0)$. Now take $a_2$ (the candies of child $1$) and get them back to child $2$ and split the candies of child $2$ in $a_1-1$ for child $1$ and $a_3+1$ for child $3$, so $(a_2,a_1+a_3,0)\rightarrow (a_1-1,a_2,a_3+1)$. By these two steps we have arrived from $(a_1,a_2,a_3,a_4,....,a_n)$ to $(a_1-1,a_2,a_3+1,a_4,...,a_n)$. This means that whenever we have $a_i\ge 1$, we can make either $a_{i-2}$ or $a_{i+2}$ bigger with $1$ by taking $1$ from $a_i$. By going inductively and using that $n$ is odd, we get that $$(a_1,a_2,...,a_n)\rightarrow (a_1-1,a_2,a_3+1,a_4,a_5,...,a_n)\rightarrow (a_1-1,a_2,a_3,a_4,a_5+1,...,a_n)\rightarrow ... (a_1-1,a_2,a_3,...,a_{n-2},a_{n-1},a_n+1)\rightarrow (a_1-1,a_2+1,a_3,...,a_n)$$so whenever $a_i\ge 1$ we can make either $a_{i-1}$ or $a_{i+1}$ bigger with $1$ by taking $1$ from $a_i$ and leaving everything else unaffected. This means that we can shift units as we wish and this readily implies that all configurations are attainable. The case when $n$ is even is a little bit more involved, but has the same underlying ideas nevertheless. First note that it is not possible to attain a configuration with $a_i=0$ for all even $i$ or for all odd $i$. Suppose that it is possible and let $(x_1,x_2,..,x_n)$ be the first configuration in which it happens. Suppose wlog that $x_i=0$ for all even $i$. Let $(y_1,...,y_n)$ be a configuration from which $(x_1,...,x_n)$ can be attained after a step. If there was $y_j\ge 1$ with odd $j$ then, by the way actions are defined, $x_{j-1}+x_{j+1}\ge 1$, which is a contradiction as $x_{j-1}=x_{j+1}=0$. So $y_j=0$ for all odd $j$, in contradiction with the the fact that $(x_1,...,x_n)$ was the first configuration to appear with $0$s on either all even or all odd positions. We can again take $1$ from $a_i\ge 1$ and give it to $a_{i+2}$ (or $a_{i-2}$), by $(a_1,a_2,a_3,a_4)\rightarrow (0,a_1+a_3,a_2+a_4,0)\rightarrow (a_1-1,a_2,a_3+1,a_4)$ and pairing everything else like in case $n$ odd. Furthermore, if $a_i\ge 2$, we can take $1$ from $a_i$ and give it to $a_{i+1}$ (or $a_{i-1}$) leaving everything else unaffected: $$(a_1,a_2,a_3,a_4)\rightarrow (a_1-1,a_3+1,a_2+a_4,0)\rightarrow (a_1-1,a_2+1,a_3,a_4)$$and pairing everything else like before. To explain better, in the first step we take $1$ from child $1$ and give it to child $2$, move all candies of children $2$ and $4$ to child $3$ and move all candies of child $3$ to child $2$. In the second step, take $1$ from child $1$ and give it to child $2$, give $1$ candy from child $2$ to child $1$ and give the other $a_3$ candy to child $3$ and from child $3$ give $a_2$ candies to child $2$ and $a_4$ candies to child $4$. By the first observation (the one with taking from $a_i$ and giving to $a_{i\pm 2}$), units move freely within even positions and they also move freely within odd positions. So we can suppose that child $1$ and child $2$ have all the candies. By the the second observation (the one with taking from $a_i$ and giving to $a_{i\pm 1}$ when $a_i\ge 2)$, we can attain all pairs $(a_1,a_2)$ such that $a_1+a_2=n,a_1,a_2>0$. Now the problem is finished: take a configuration which has a total of $a>0$ candies on odd positions and $b>0$ candies on even positions, $a+b=n$. Then, by the previous, we can obtain the configuration $(a,b,0,0,...,0)$ and as units move freely withing even, respectively within odd positions, we get that the configuration is clearly attainable.