B. Given that $a$ and $b$ are positive integers, and $b-a=2013$. If the equation $x^{2}-a x+b=0$ has positive integer solutions, then the minimum value of $a$ is $\qquad$ .
B. 93.

Let the two solutions of the original equation be $m$ and $n$. Then
$$
m+n=a, \quad mn=b \text{.}
$$

Assume without loss of generality that $m$ is a positive integer. Then $n=a-m$ is an integer, and $n=\frac{b}{m}>0$.
Therefore, $n$ is also a positive integer.
Clearly, 1 is not a solution to the original equation.
Thus, $m-1$ and $n-1$ are also positive integers.
$$
\begin{array}{l}
\text{By } (m-1)(n-1) = mn - (m+n) + 1 \\
= b - a + 1 = 2014,
\end{array}
$$

Since $2014 = 2 \times 19 \times 53$, we know that when $m-1$ and $n-1$ are 38 and 53 respectively, their sum is the smallest.
Notice that,
$$
a = m + n = (m-1) + (n-1) + 2 \text{.}
$$

Therefore, the minimum value of $a$ is $38 + 53 + 2 = 93$.