Minimal number of "words" that contain all possible pairs of letters in all position pairs Defining a word as sequence of ordered letters ($1$..$q$ letters) of length L, 
what is the minimal number of words such that among the entire list of words,
at every pair of positions, I can find any two letters?
for example, for $q=3$ and $L=2$
here is the minimal list:
$$1 1,
2 2,
3 3,
1 2,
2 3,
3 1,
1 3,
2 1,
3 2,$$
total $q^2$ words are needed.
but for $L=3$ the minimal number is still $q^2$, obtained by:
$$1 1 1,
2 2 1,
3 3 1,
1 2 2,
2 3 2,
3 1 2,
1 3 3,
2 1 3,
3 2 3,$$
for $L=4$ the number is different...
what is the minimal number of words for $(q,L)$ and specifically, what is the asymptotic value for $L\gg 1$?
Thanks for you answers!
 A: A Possible Solution: Actually given your family for $q=L=3,$ we can duplicate it (side by side) with the goal of achieving the goal for $L=6.$ Here, all $(i,j)$ with $i\leq 3,j\geq 4,\quad i\neq j\pmod 3$ positions will also achieve all 2-tuples by design.
So you can insert extra rows at the bottom with x incidating don't care to cover those $i,j$ in the left and right halves with $i=j \pmod 3.$
$$
\begin{array}{c}
111~111\\
221~221\\
331~331\\
122~122\\
232~232\\
312~312\\
133~133\\
213~213\\
321~321\\ 
 \\
111~222\\
222~111\\
111~333\\
333~111\\
222~333\\
333~222\\
\end{array}
$$
If I'm not missing something this gives a gain in efficiency.
Now recursively double the new full array, this will then require you covering some other pairs whose indices are the same mod 6.
Earlier Discussion:
What you may use is an orthogonal array of strength  $=2$ and repetition $\lambda=1$ over a
$q−$ary alphabet. This is an array of symbols from the alphabet, where looking down any two columns each possible pair of symbols appear once ($\lambda=1)$.
The exactly once property is not necessarily needed for your purposes, but it yields constructions that are uniform with respect to the symbol pair positions. This may be natural feature of an optimal solution.
Hedayat and Sloane have a nice book on orthogonal arrays. The talk at the Isaac Newton institute, video and slides available here  is a nice overview.
Rao’s and Bose-Bush’s bounds apply generally, but can be improved in special cases.
From page 10 of the slides, the general lower bound (translated into your variables with $N$ the number of rows)
$$
N\geq L(q-1)+1
$$
can be obtained. Strengthening this in general is hard since the standard results rely on the condition $\lambda-1$ being nonzero modulo some parameter but your $\lambda=1.$
A: First, let me elaborate the upper bound from @kodlu’s answer. If $q$ is a power of a prime, then $q^2$ words suffice for $L=q+1$ (for $L=q$ it is almost trivial, improving by $1$ needs just a bit more). Then, doubling $L$ increases the number of words by $q(q-1)$, so for those values of $L$ it suffices to take $q(q-1)\log_2\frac L{q+1}+q^2$ words.
Let me show a somewhat close lower bound. Let $w$ be the number of words; set $k=w-q(q-1)+1$. Take any $k$ words. Let $v_i$ be a vector composed from all $i$th entries of the $k$ words. If two of those vectors, say $v_i$ and $v_j$, coincide, this means that at most $w-k=q(q-1)-1$ words differ in positions $i$ and $j$, so not all pairs are covered.
Thus, we have $L$ distinct vectors in $[q]^k$, so $L\leq q^k$, or $w\geq q(q-1)+1+\log_qL$.
Therefore, the growth rate is indeed logarithmic (but the constant at the logarithm is yet unclear).
Addendum. Let me present an example for $L=q+1$, when $q$ is a power of a prime.
Consider an affine plane $\mathbb F_q^2$. All lines in it are partitioned into $q+1$ classes $C_1,\dots,C_{q+1}$ of mutually parallel lines (one class consists of all lines with equations of the form $ax+by=c$ with fixed $(a:b)$). Enumerate the lines in each class by numbers from 1 to $q$.
For each point $p\in \mathbb F_q^2$, take a word $w_1\dots w_{q+1}$ where $w_i$ is the number of the line in $C_i$ passing through $p$. Then, for any two classes $C_i$ and $C_j$ and for any two lines in them, the lines meet at a unique point, which means exactly what we need to get in the $i$th and $j$th positions.
