Dirichlet's Box Principle

Theorem
Let $S$ be a finite set whose cardinality is $n$.

Let $S_1, S_2, \ldots, S_k$ be a partition of $S$ into $k$ subsets.

Then at least one subset $S_i$ of $S$ contains at least $\left \lceil {n/k} \right \rceil$ elements.

It can also be seen in the simpler form:

If a set of $n$ distinct objects is partitioned into $k$ subsets, where $0 < k < n$, then at least one subset must contain at least two elements.

Proof
Suppose this were not the case, and no subset $S_i$ of $S$ has as many as $\left \lceil {n/k} \right \rceil$ elements.

Then the maximum number of elements of any $S_i$ would be $\left \lceil {n/k} \right \rceil - 1$.

So the total number of elements of $S$ would be no more than $k \left({\left \lceil {n/k} \right \rceil - 1}\right) = k \left \lceil {n/k} \right \rceil - k$.

There are two cases:
 * $n$ is divisible by $k$;
 * $n$ is not divisible by $k$.

Suppose $k \backslash n$.

Then $\left \lceil {\dfrac n k} \right \rceil = \dfrac n k$ is an integer and $k \left \lceil {n/k} \right \rceil - k = n - k$.

Thus $\displaystyle \sum_{i=1}^k \left \vert {S_i}\right \vert \le n-k < n$.

This contradicts our assumption that no subset $S_i$ of $S$ has as many as $\left \lceil {n/k} \right \rceil$ elements.

Next, suppose that $k \nmid n$. Then $k \left \lceil {n/k} \right \rceil - k < \dfrac {k \left({n+1}\right) - k}{k} = n$ and again this contradicts our assumption that no subset $S_i$ of $S$ has as many as $\left \lceil {n/k} \right \rceil$ elements.

Either way, there has to be at least $\left \lceil {n/k} \right \rceil$ elements in at least one $S_i \subseteq S$.

Historical Note
This principle has been attributed to Dirichlet.