A funny twist on Hall's Marriage Theorem: in an unsatisfiable set of preferences, there must be a set $A$ of students the union of whose acceptable gifts is of size less than $|A|$. Then the contribution to the given sum from students in $A$ is at least $\frac{|A|}{|A|-1}>1$.

It seems it is really easy: as some number $x_i \ge n$ it means santa will always have the possibility to mahe that child happy, regardless of the other numbers, now the same process can be repeated with $n-1$ numbers $x_i$. Hence solved as each time we have one child we can make happy. Edit: this idea can be used to prove JBL s remark with induction, this because $\frac x {x-1}$ is a decreasing sequence.

Consider the bipartite graph in a natural fashion, and label each $n$ vertices for the student group with the number $x_i$ to denote its degree. Take a subset $S$ of the $n$ vertices with $s < n$ elements, let they be $i, i+1, ..., i+s-1$. We have, since $x_i \le n$, then $\sum_{p=i}^{i+s-1} \dfrac{1}{x_p} < 1$. Let $k$ be a vertex in $S$ with maximal degree, and suppose $x_k \le s \implies \dfrac{1}{x_k} \ge \dfrac{1}{s}$. Then, $1 > \sum_{p=i}^{i+s-1} \dfrac{1}{x_p} \ge s \cdot \dfrac{1}{x_k} \ge s \cdot \dfrac{1}{s} \ge 1$ which is a contradiction $\implies x_k > s$. Hence, for any subset $S$ with $s < n$ elements, thee number of adjacent vertices will be at least $s+1$. For $s=n$, we can achieve equality, and the argument holds as $x_k < n \implies 1 \ge \sum_{p=1}^n \dfrac{1}{x_p} \ge n \cdot \dfrac{1}{n} > 1$. Hence by Hall's Marriage theorem, there exists a perfect matching, which is what we wanted to prove.

We note that for each subset $S$ of $\{1,2,...,n\}$, we have $\displaystyle\sum_{i\in S}\frac{1}{x_i}\le 1$. By Cauchy we get $\displaystyle\sum_{i\in S}{x_i}\ge |S|^2$. But every present can only be preferred by at most $|S|$ children from $S$, so it is clear that the set of presents that children from $S$ prefers is at least of size $|S|$. So by Hall's Marriage we are done.

Deleted wrong post.

It is at least n gifts

Thank you. I was actually looking for a qualifier like that. I don't know why I did not see it.

Nice and easy. By Hall's Marriage Lemma we conclude if $x_{i_1} +...x_{i_k} \ge k \forall k\le n$, ${i_1,...,i_k}$ is any $k$ subset of ${1,...,n}$. Otherwise this subset contributes at least $\frac{k}{k-1}$ to the sum which is greater than $1$.

Solved with Philip Hall kids are gifts and gifts are boxes.

Suppose that there exist a set $S$ of $k\leq n$ children (identified by their numbers, so $S$ is a set of integers) such that there are at most $k-1$ gifts who at least one of the children want. Then $$\frac{1}{x_1}+\cdots+\frac{1}{x_n}\geq \sum_{i \in S} \frac{1}{x_i}>\sum_{i \in S} \frac{1}{k}=1,$$contradiction. Thus Hall's Condition is satisfied so a perfect matching exists. $\blacksquare$

Observe that for any $k\leq n$ children, we still have that $\frac{1}{x_{a_{i_1}}}+ \frac{1}{x_{a_{i_2}}}+...+\frac{1}{x_{a_{i_k}}} \leq 1$. By Cauchy Schwarz Inequality, $$ (x_{a_{i_1}}+x_{a_{i_2}}+...+x_{a_{i_k}})(\frac{1}{x_{a_{i_1}}}+ \frac{1}{x_{a_{i_2}}}+...+\frac{1}{x_{a_{i_k}}}) \geq k^2$$and since $\frac{1}{x_{a_{i_1}}}+ \frac{1}{x_{a_{i_2}}}+...+\frac{1}{x_{a_{i_k}}} \leq 1$, we have that $(x_{a_{i_1}}+x_{a_{i_2}}+...+x_{a_{i_k}}) \geq k^2$, so $\max_{1 \leq j \leq k} a_{i_j} \geq k$, so the neighborhood of any subset of $k \leq n$ children has size at least $k$. Therefore, Hall's condition is satisfied, so by Hall's Marriage Theorem, each child can receive a desired gift. $\blacksquare$