4. Write 2013 different real numbers on 2013 cards. Place the cards face down on the table. Two players, A and B, play the following game: in each round, A can arbitrarily select ten cards, and B will tell A one of the ten numbers written on these cards (B does not tell A which card the number is written on). Find the maximum positive integer $t$, such that A can certainly determine the numbers on $t$ cards after a finite number of rounds.
4. The maximum value of $t$ is $1986=2013-27$.

Let $A_{1}, A_{2}, \cdots, A_{2013}$ be these 2013 cards.
First, note that player B has a strategy to prevent player A from determining the number written on any of the cards $A_{1}, A_{2}, \cdots, A_{27}$.
B divides $T=\{1,2, \cdots, 27\}$ into nine groups
$T_{i}=\{3 i-2,3 i-1,3 i\}(1 \leqslant i \leqslant 9)$.
For any ten-element subset $B$ of $A=\{1,2, \cdots, 2013\}$, consider the following two cases.
(1) If $B \backslash T \neq \varnothing$, let
$i_{0}=\min \{i \mid i \in B \backslash T\}$,

then B tells A the number on $A_{i_{0}}$.
(2) If $B \backslash T=\varnothing$, let
$i_{0}=\min \left\{i\left|T_{i} \cap B\right| \geqslant 2\right\}$,

then consider the following two sub-cases.
(i) If $\left|T_{i_{0}} \cap B\right|=3$, then B tells A the number on $A_{3 i_{0}}$;
(ii) If $\left|T_{i_{0}} \cap B\right|=2$, then B has two options to tell A.
For $T_{i_{0}} \cap B$
$=\{3 i-2,3 i-1\},\{3 i-1,3 i\},\{3 i, 3 i-2\}$.
【Option 1】B tells A the numbers on $A_{3 i-2}, A_{3 i-1}, A_{3 i}$ respectively.

【Option 2】B tells A the numbers on $A_{3 i-1}, A_{3 i}, A_{3 i-2}$ respectively.

Since A does not know which option B uses, A cannot determine the number on any of the first 27 cards (in fact, if the number on $A_{3 i-2}$ is rewritten to $A_{3 i-1}$, the number on $A_{3 i-1}$ is rewritten to $A_{3 i}$, and the number on $A_{3 i}$ is rewritten to $A_{3 i-2}$, then for any ten cards specified by A, B's response for the rewritten Option 2 is exactly the same as for the original Option 1).

Next, we show that for any 28 cards, A has a way to determine the number on one of them.
To do this, we first prove a lemma in graph theory.
Lemma $n(n \geqslant 2)$ is a positive integer. In a graph with at least $3 n-2$ vertices and at most $3 n-2$ edges, there must exist $n$ vertices such that no two of them are connected by an edge.
Proof By induction on $n$.
Assume the graph has exactly $3 n-2$ vertices and $3 n-2$ edges.

When $n=2$, if the number of edges is less than $\mathrm{C}_{4}^{2}=6$, then there are two vertices not connected by an edge.

Assume $n>2$. Let $d_{1}, d_{2}, \cdots, d_{3 n-2}$ be the degrees of the vertices. Then
$d_{1}+d_{2}+\cdots+d_{3 n-2}=2(3 n-2)$.
This indicates that either all $d_{i}=2$ or there exist $i, j$ such that
$d_{i}<2<d_{j}$.
In the first case, remove any vertex (e.g., the $i$-th) and its two adjacent vertices; in the second case, remove the $i$-th and $j$-th vertices and the vertex adjacent to the $i$-th vertex. This removes at most three vertices and at least three edges. Apply the induction hypothesis to the remaining graph to get $n-1$ vertices not connected by any edge, and add the $i$-th vertex to get $n$ vertices not connected by any edge.
The lemma is proved.
Let $A_{1}, A_{2}, \cdots, A_{28}$ be 28 cards.
$M$ denotes the set of all ten-element subsets of $\left\{A_{1}, A_{2}, \cdots, A_{28}\right\}$.

Define a function $f$ on $M$: for any $B \in M, f(B)$ represents the number B tells A for $B$. Let $\left\{c_{1}, c_{2}, \cdots, c_{k}\right\}$ $(k \leqslant 28)$ be the range of $f$.
Let $S_{i}=\bigcap_{B \in f^{-1}\left(c_{i}\right)} B$.
If for some $i, S_{i}=\left\{A_{m}\right\}$ is a singleton, then the number on $A_{m}$ is $c_{i}$.

If for any $i(1 \leqslant i \leqslant k), S_{i}$ contains at least two elements, select two elements from each $S_{i}$ and connect them with an edge, forming a graph with 28 vertices and no more than 28 edges. According to the lemma $(n=10)$, there exist 10 vertices in the graph that are not connected by any edge.

Let the set of these 10 vertices be $B \in M$, $f(B)=c_{i}$. Then $S_{i} \subset B$. $S_{i}$ contains at least two vertices connected by an edge, i.e., there are two vertices in $B$ connected by an edge, which is a contradiction.