4. Let $s=\sum_{k=1}^{2015} k 2^{k}$. Then the remainder when $s$ is divided by 100 is
4.6.
$$
\begin{array}{l}
\text { Given } s=\sum_{k=1}^{2015} k 2^{k} \\
\Rightarrow 2 s=\sum_{k=1}^{2015} k 2^{k+1}=\sum_{k=1}^{2016}(k-1) 2^{k} \\
\Rightarrow s=2 s-s=\sum_{k=1}^{2016}(k-1) 2^{k}-\sum_{k=1}^{2015} k 2^{k} \\
=2015 \times 2^{2016}-\sum_{k=1}^{2015} 2^{k} \\
=2014 \times 2^{2016}+2 \text {. } \\
\end{array}
$$

Given $2^{10}=1024$, we know that for any positive integer $n>1$,
$$
2^{20+n}-2^{n}=2^{n} \times 1025 \times 1023
$$

is a multiple of 100.
Thus, $2^{2016}$ modulo 100 is the same as $2^{16}$ modulo 100, both of which are 36. Therefore, the remainder when $s$ is divided by 100 is 6.