  Suppose $G$ is a connected simple graph with the vertex set $V( G ) = \{
v_1,v_2,\cdots ,v_n \} $. Let $d_G( v_i,v_j ) $ be the least distance between
$v_i$ and $v_j$ in $G$. Then the distance matrix of $G$ is $D( G ) =( d_{ij} )
_{n\times n}$, where $d_{ij}=d_G( v_i,v_j ) $. Since $D( G )$ is a non-negative
real symmetric matrix, its eigenvalues can be arranged as $\lambda_1(G)\ge
\lambda_2(G)\ge \cdots \ge \lambda_n(G)$, where eigenvalue $\lambda_n(G)$ is
called the least distance eigenvalue of $G$. In this paper we determine the
unique graph whose least distance eigenvalue attains maximum among all
complements of graphs of diameter greater than three.
