Fictional Lockers

We will begin by observing the events.

• After the first student operates, all lockers are opened.

• After the second student operates, lockers with numbers $2,4,6,8,.........,200$ are closed again.

• After the third student operates, lockers with numbers $3,9,15,21,.....,195$ are closed and lockers with numbers $6,12,18,24.......,198$ are opened again.

So we observe that the lockers with locker no.s having odd positive factors remain open, and those having even factors remain closed after all operations.

So we need to find no.s with odd factors.

If the no. $k$ can be represented as $p_{1}^{n_{1}}\times p_{2}^{n_{2}}\times p_{3}^{n_{3}}\times ........................$

then the no. of factors is $(n_1+1)\times (n_2+1)\times(n_3+1)\times ........................$ which will be odd when $(n_1+1),(n_2+1),(n_3+1)....$ all are odd.

Thus $n_1,n_2,n_3.....$ all should be even.

If $n_1,n_2,n_3.....$ are all even then $k$ must be perfect square.

So, we conclude that all lockers with locker no.s a perfect squares are all open after all operation.

And there are $14$ perfect squares till the no. $200$ . So the answer is $\boxed {14}$