There are two different cases.
If the observed values differ by $1$, this implies known values for $X_3$ and $X_4$, and the lower of the two observed values is the value of $X_1+X_2$. If this is $0$ or $2$, it implies the value $0$ or $1$, respectively, for $X_1$; if it is $1$, both values of $X_1$ are equally likely.
On the other hand, if the observed values are equal, this implies that the values of $X_3$ and $X_4$ are equal. Given that they are equal, the problem effectively becomes finding the maximum likelihood estimate for $X_1$ given $X_1+X_2+X_3$. If the observed value is $0$ or $3$, that determines $X_1$ with value $0$ or $1$, respectively. If the observed value is $1$ or $2$, then the value of $X_1$ is $0$ or $1$, respectively, with probability $2/3$, and $1$ or $0$, respectively, with probability $1/3$, so the maximum likelihood estimate in this case is $0$ or $1$, respectively.
In summary, the maximum likelihood estimate for $X_1$ given the two observed values is given by
$ \begin{array}{c|cccc} &0&1&2&3\\\hline 0&0&0&-&-\\ 1&0&0&=&-\\ 2&-&=&1&1&\\ 3&-&-&1&1 \end{array} $
where $-$ indicates a case that cannot occur and $=$ indicates that both values of $X_1$ are equally likely.