1. Among the numbers $1,2, \cdots, 100$, add “+” and “ _ ” signs so that the algebraic sum of the resulting expression is 4150. The maximum number of “+” signs that can be added is ( ) .
(A) 92
(B) 93
(C) 94
(D) 95
- .1.C.

Let the sum of all positive numbers be $x$, and the sum of all negative numbers be $y$. Then
$$
\left\{\begin{array} { l } 
{ x + y = 4150 , } \\
{ x + | y | = \frac { 100 \times ( 1 + 100 ) } { 2 } , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
x+y=4150, \\
x-y=5050 .
\end{array}\right.\right.
$$

Solving, we get $y=-450$.
There are 99 plus and minus signs to be added. To maximize the number of plus signs, we need to minimize the number of minus signs. Since
$$
-100-99-98-97-56=-450 \text {, }
$$

the minimum number of minus signs required is 5.
Therefore, the maximum number of plus signs that can be added is $99-5=94$.