Just to clarify, when you wrote sequence does it mean something that has a common difference or common ratio or just any jumbled up order of some integers?

The latter. @below, it's not specified in the Russian text, but I suppose the numbers in the sequence are distinct.

Must the sequence that the spectator writes be distinct numbers or could it possibly be (1,1,...,1)?

The spectator writes "sequence" not set. So, the numbers are ordered and not necessaily distinct, I think. I assume the problem statement is: Spectator writes $a = (a_1, a_2, \ldots, a_{10})$ and assistant add $b(a) = (b_1, b_2, ..., b_k)$. Now, $s = (a_1, a_2, \ldots, a_{10}, b_1, b_2, \ldots, b_k)$ are written in the black board. Spectator replaces at most three place of $s$ and let $r$ be the resulting sequence. Magician sees $r = (r_1, r_2, \ldots, r_{10+k})$ and he must guess original $a$.

This problem is almost same as block code https://en.wikipedia.org/wiki/Block_code of $(k+10, 10, 7)_{239}$ (See the popular notation) which means sending message with length 10 adding $k$ additional message and hamming distance of any two resulting messages (which length are $k+10$) must be at least 7. From the singleton bound, we can show $6\leq k$. This is not so hard to show without the theory. But constructing the case for $k=6$ is very hard. At least we need some theory like Reed-Solomon error correction. https://en.wikipedia.org/wiki/Reed%E2%80%93Solomon_error_correction In my opinion, this problem is unsuitable for olympiad problem for high school students because the contestant who knows the theory is advantageous to solve it while others are very hard to solve.

toshihiro shimizu wrote: In my opinion, this problem is unsuitable for olympiad problem for high school students because the contestant who knows the theory is advantageous to solve it while others are very hard to solve. It's not hard. Let the helper write polynomial $f\in Z_{239}[t]$ of degree $\leq 9$ for which $f(i)=x_i$ at $1\leq i\leq 10$, and add the numbers $f(11),\dots, f(16)$ to the right-hand side. The spectator will change three numbers, the magician will comes in and tries all the supposed sets of 13 unchanged numbers. Each he will interpolate the corresponding points with a polynomial of degree $\leq 12$. In one case (when he guessed the 13 numbers correctly) the degree is actually $\leq 9$, and putting the points from $1$ to $\leq 9$ back into this the points from 1 to 10 into this polynomial again, the magician will know the original numbers. The two two different polynomials of degree $\leq 9$ cannot appear. In fact. each polynomial is the same as the original polynomial at all points except for three points, so two different polynomials coincide with each other on all points points except for six, i.e., at $10$ points.