[sp]
It is possible if and only if $n\equiv0\text{ or }1\pmod4$.
To prove that the condition is necessary, we note that each switch must be flicked an odd number of times. As the total number of flicks is $\dfrac{n(n+1)}{2}$, we must have:
$$\begin{align*}
n &\equiv\frac{n(n+1)}{2}\quad&\pmod2\\
2n &\equiv n(n+1)&\pmod4\\
n(n-1)&\equiv0 &\pmod4
\end{align*}$$
and, as $n$ and $n-1$ cannot both be even, the conclusion follows.
To prove that the condition is sufficient, we proceed by induction with a step of 4. The base cases are $n=1$ and $n=4$. The first case is obvious, and, for the second case, we have the solution:
$$
\begin{array}{c|c|c|c|c|}
&1&2&3&4\\
\hline
1&1&0&0&0\\
2&0&1&1&0\\
3&1&1&1&0\\
4&1&1&1&1\\
\hline
\text{total}&3&3&3&1
\end{array}
$$
where '1' stands for a flick, each column after the first corresponds to a switch and each row corresponds to a round.
We now assume that there is a solution for $n=k$ and show that we can extend it to a solution for $n=k+4$. We add 4 columns and 4 rows to the table.
The first $k$ rows are extended with 0. In the rows $k+1$ to $k+4$, we fill the first $k$ columns with 1, and the last 4 columns with a copy of the solution for $n=4$. This is indeed a solution, because:
- Row $i$ contains $i$ 1's.
- The parity of the first $k$ column totals has not changed; it is odd by the induction hypothesis.
- The parity of the last 4 column totals is odd as it is in the case $n=4$.
The solutions for $n=5$ and $n=8$ are shown below:
$$
\begin{array}[t]{c|c|c|c|c|c|}
&1&2&3&4&5\\
\hline
1&1&0&0&0&0\\
2&1&1&0&0&0\\
3&1&0&1&1&0\\
4&1&1&1&1&0\\
5&1&1&1&1&1\\
\hline
\text{total}&5&3&3&3&1
\end{array}
\qquad\begin{array}[t]{c|c|c|c|c|c|c|c|c|}
&1&2&3&4&5&6&7&8\\
\hline
1&1&0&0&0&0&0&0&0\\
2&0&1&1&0&0&0&0&0\\
3&1&1&1&0&0&0&0&0\\
4&1&1&1&1&0&0&0&0\\
5&1&1&1&1&1&0&0&0\\
6&1&1&1&1&0&1&1&0\\
7&1&1&1&1&1&1&1&0\\
8&1&1&1&1&1&1&1&1\\
\hline
\text{total}&7&7&7&5&3&3&3&1
\end{array}
$$
[/sp]