Here is what I remember from a proof I came up with long time ago (it appeared in some competitions). I am sure it is known, but since the proof is short, I will put it here:

The statement can be reduced to the case $n=p$ is prime. Now it will follow from the following:

Lemma: Let $2\leq i\leq p$ and consider any set $A$ of elements $ a_1,\cdots, a_{2i-1} \in \mathbb F_p$ such that no $i$ of them are the same. Let $S$ be the set of all sums of $i$ elements of $A$. Then $|S|\geq i$.

Proof: Induction on $i$, the case $i=2$ is easy. Suppose the result is true for $p>i=k\geq 2$. Consider the set $A'$ of $2k+1$ elements $\{a_1,\cdots,a_{2k-1},b,c\}$, WLOG we may assume that $b,c$ represent the two most frequently appeared elements in $A'$. By assumption $b\neq c$.

The set $\{a_1,\cdots,a_{2k-1}\}$ satisfies the condition that any frequency is less than $k$, otherwise there will be $3$ elements appearing at least $k$ times in $A'$, impossible. By induction we can choose $k$ subsets of $k$ elements with sums are distinct $s_1,\cdots, s_{k}\in \mathbb F_p$. Let $B=\{b+s_1,\cdots, b+s_{k}\}$ and $C=\{c+s_1,\cdots, c+s_{k}\}$, obviously $|B|=|C|=k$. We will be done by showing $B \neq C$. If so, then by summing all the elements in each set one gets $k(b-c)=0$, impossible. QED.

Applying the lemma for $i=p$, note that if there are $p$ identical elements then their sum is $0$.

By the way, I asked some question on generalizations of this theorem, and get really good answers here.

Here is what I remember from a proof I came up with long time ago (it appeared in some competitions). I am sure it is known, but since the proof is short, I will put it here:

The statement can be reduced to the case when $n=p$ is prime. Now it will follow from the following:

Lemma: Let $2\leq i\leq p$ and consider any set $A$ of elements $ a_1,\cdots, a_{2i-1} \in \mathbb F_p$ such that no $i$ of them are the same. Let $S$ be the set of all sums of $i$ elements of $A$. Then $\left|S\right|\geq i$.

Proof: Induction on $i$, the case $i=2$ is easy. Suppose the result is true for $p>i=k\geq 2$. Consider the set $A'$ of $2k+1$ elements $\{a_1,\cdots,a_{2k-1},b,c\}$, WLOG we may assume that $b,c$ represent the two most frequently appearing elements in $A'$. By assumption $b\neq c$.

The set $\{a_1,\cdots,a_{2k-1}\}$ satisfies the condition that any frequency is less than $k$, otherwise there will be $3$ elements appearing at least $k$ times in $A'$, impossible. By induction we can choose $k$ subsets of $k$ elements whose sums $s_1,\cdots, s_{k}\in \mathbb F_p$ are distinct. Let $B=\{b+s_1,\cdots, b+s_{k}\}$ and $C=\{c+s_1,\cdots, c+s_{k}\}$, obviously $\left|B\right|=\left|C\right|=k$. We will be done by showing $B \neq C$. If not, then by summing all the elements in each set one gets $k(b-c)=0$, impossible. QED.

Applying the lemma for $i=p$, note that if there are $p$ identical elements then their sum is $0$.

By the way, I asked some question on generalizations of this theorem, and get really good answers here.

Here is what I remember from a proof I came up with long time ago (it appeared in some competitions). I am sure it is known, but since the proof is short, I will put it here:

The statement can be reduced to the case $n=p$ is prime. Now it will follow from the following:

Lemma: Let $2\leq i\leq p$ and consider any set $A$ of elements $ a_1,\cdots, a_{2i-1} \in \mathbb F_p$ such that no $i$ of them are the same. Let $S$ be the set of all sums of $i$ elements of $A$. Then $|S|\geq i$.

Proof: Induction on $i$, the case $i=2$ is easy. Suppose the result is true for $p>i=k\geq 2$. Consider the set $A'$ of $2k+1$ elements $\{a_1,\cdots,a_{2k-1},b,c\}$, WLOG we may assume that $b,c$ represent the two most frequently appeared elements in $A'$. By assumption $b\neq c$.

The set $\{a_1,\cdots,a_{2k-1}\}$ satisfies the condition that any frequency is less than $k$, otherwise there will be $3$ elements appearing at least $k$ times in $A'$, impossible. By induction we can choose $k$ subsets of $k$ elements with sums are distinct $s_1,\cdots, s_{k}\in \mathbb F_p$. Let $B=\{b+s_1,\cdots, b+s_{k}\}$ and $C=\{c+s_1,\cdots, c+s_{k}\}$, obviously $|B|=|C|=k$. We will be done by showing $B \neq C$. If so, then by summing all the elements in each set one gets $k(b-c)=0$, impossible. QED.

Applying the lemma for $i=p$, note that if there are $p$ identical elements then their sum is $0$.

By the way, I asked some question on generalizations of this theorem, and get really good answers here.

Here is what I remember from a proof I came up with long time ago (it appeared in some competitions). I am sure it is known, but since the proof is short, I will put it here:

The statement can be reduced to the case $n=p$ is prime. Now it will follow from the following:

Lemma: Let $2\leq i\leq p$ and consider any set $A$ of elements $ a_1,\cdots, a_{2i-1} \in \mathbb F_p$ such that no $i$ of them are the same. Let $S$ be the set of all sums of $i$ elements of $A$. Then $|S|\geq i$.

Proof: Induction on $i$, the case $i=2$ is easy. Suppose the result is true for $p>i=k\geq 2$. Consider the set $A'$ of $2k+1$ elements $\{a_1,\cdots,a_{2k-1},b,c\}$, WLOG we may assume that $b,c$ represent the two most frequently appeared elements in $A'$. By assumption $b\neq c$.

The set $\{a_1,\cdots,a_{2k-1}\}$ satisfies the condition that any frequency is less than $k$, otherwise there will be $3$ elements appearing at least $k$ times in $A'$, impossible. By induction we can choose $k$ subsets of $k$ elements with sums are distinct $s_1,\cdots, s_{k}\in \mathbb F_p$. Let $B=\{b+s_1,\cdots, b+s_{k}\}$ and $C=\{c+s_1,\cdots, c+s_{k}\}$, obviously $|B|=|C|=k$. We will be done by showing $B \neq C$. If so, then by summing all the elements in each set one gets $k(b-c)=0$, impossible. QED.

Applying the lemma for $i=p$, note that if there are $p$ identical elements then their sum is $0$.

By the way, I asked some question on generalizations of this theorem, and get really good answers here.