jmc

The sum of the arithmetic series $1+2+3+\cdots +n$ is equal to $\frac{n(n+1)}{2}$. Let $k$ and $k+1$ be the two consecutive integers removed, so that their sum is $2k+1$. It follows that $$\frac{n(n+1)}{2}-(2k+1)=241.$$

The smallest numbers that Charlize could have omitted are 1 and 2, so $$241=\frac{n(n+1)}{2}-(2k+1)\le \frac{n(n+1)}{2}-3,$$ which gives us the inequality $n(n+1)\ge 488$. If $n=21$, then $n(n+1)=462$, and if $n=22$, then $n(n+1)=506$, so $n$ must be at least 22.

The largest numbers that Charlize could have omitted are $n$ and $n-1$, so $$241=\frac{n(n+1)}{2}-(2k+1)\ge \frac{n(n+1)}{2}-n-(n-1)=\frac{(n-1)(n-2)}{2},$$ which gives us the inequality $(n-1)(n-2)\le 482$. If $n=23$, then $(n-1)(n-2)=462$, and if $n=24$, then $(n-1)(n-2)=506$, so $n$ must be at most 23.

From the bounds above, we see that the only possible values of $n$ are 22 and 23.

If $n=22$, then the equation $$\frac{n(n+1)}{2}-(2k+1)=241$$ becomes $253-(2k+1)=241$, so $2k+1=12$. This is impossible, because $2k+1$ must be an odd integer.

Therefore, $n=\boxed{23}$. Note that $n=23$ is possible, because Charlize can omit the numbers 17 and 18 to get the sum $23\cdot 24/2-17-18=241$.