3. If $\left\{a_{n}\right\}$ is an arithmetic sequence, the first term $a_{1}>0$, and
$$
a_{2013}+a_{2014}>0, a_{2013} a_{2014}<0,
$$
then the largest natural number $n$ for which $a_{n}>0$ is ( ).
(A) 4025
(B) 4026
(C) 4027
(D) 4028
3. B.

From the problem, we know that in this sequence, the terms from the 1st to the 2013th are positive, and from the 2014th term onwards, they are negative. Clearly, when \( n = 2013 \), the sum of all positive terms \( S_{n} \) reaches its maximum value.

Furthermore, \( S_{n} \) is a quadratic function of \( n \) without a constant term, and it opens downwards. Therefore, the 2013th term is closest to the axis of symmetry, which means the axis of symmetry lies between 2013 and 2013.5.

Since the graph of the quadratic function intersects the \( x \)-axis at the point \((0,0)\), let the other intersection point be \((x, 0)\). Thus, \( x \) should lie between 4026 and 4027 (as shown in Figure 3).
Therefore, the largest natural number for which \( S_{n} > 0 \) is 4026.