3. Let the integer $k$ be such that the quadratic equation in $x$
$$
2 x^{2}+(11-5 k) x+3 k^{2}-13 k-2003=0
$$

has at least one integer root. Then all such integers $k$ are
$\qquad$
3. $-4030,-2012,2018,4036$.

Completing the square, we get
$$
(x-k+2)(2 x-3 k+7)=2017 \text {. }
$$

Since 2017 is a prime number, we have
$$
\left\{\begin{array} { l } 
{ x - k + 2 = \pm 1 } \\
{ 2 x - 3 k + 7 = \pm 2 0 1 7 , }
\end{array} \text { or } \left\{\begin{array}{l}
x-k+2= \pm 2017, \\
2 x-3 k+7= \pm 1 .
\end{array}\right.\right.
$$

Solving these, we get
$$
\begin{aligned}
(x, k)= & (-2013,-2012),(6051,4036), \\
& (2015,2018),(-6049,-4030) .
\end{aligned}
$$

Therefore, the integers $k$ are $-4030,-2012,2018,4036$.