50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010)

a) If we have all $+$, then: $1+2+3+\cdots +62=1953<2010$, $1+2+3+\cdots +63=2016>2011$. Hence, $N\ge 63$. From the other hand, $N=63$ will do: $1+2-3+\cdots +62+63=2016-6=2010$.

b) $N=63,64$ are not the solutions, because change of sign before any number does not change the parity of the expression itself and $1+2+3+\cdots +63=2016$ and $1+2+3+\cdots +64$ are even. Hence, $N\ge 65$. Now it's easy to locate signs. Since $1+2+3+\cdots +65=2145$, then $-2\cdot (2+65)+1+2+3+\cdots +65=2145-134=2011$.