jmc

On the first day, $1+2=3$ students know the secret. On the second day, $1+2+4=7$ students know the secret. On the third day, $1+2+4+8=15$ students know the secret. Notice that each of these sums is one less than the next power of 2. Therefore, on the $n$th day, $1+2+\cdots +2^n=2^{n+1}-1$ students know the secret. Setting $2^{n+1}-1=1023$, we find $2^{n+1}=1024⟹n+1=10⟹n=9$. We counted Monday as the first day, so the eighth day is Monday and the ninth day is $\boxed{\text{Tuesday}}$.

Note: To show that $1+2+\cdots +2^n=2^{n+1}-1$, define the sum to be $s$ and multiply both sides of $$s=1+2+\cdots +2^n,$$by 2 to find $$2s=2+4+\cdots +2^{n+1}.$$Subtract the first equation from the second to obtain $s=2^{n+1}-1$.