Consider the polynomial $P(x) = \prod \limits_{1 \leqslant i \leqslant 2100} (1+x^i)$. We need to find the sum of coefficients of $x^u$ for all $u \equiv 3 \pmod 7$. Let $\omega = e^{\frac{2\pi i}{7}}$ and observe that then this sum is basically just \[\frac{1}{7}\sum_{0 \leqslant k \leqslant 6} \omega^{4k}P(\omega^k)\] For any $7$th root of unity $\zeta$, $P(x) = \left(\prod \limits_{0 \leqslant i \leqslant 6} (1+\zeta^i)\right)^{300}$. For $\zeta \ne 1$, this quantity is fixed, call it $C$. Our sum is now \[\frac{1}{7}\sum_{0 \leqslant k \leqslant 6} \omega^{4k}P(\omega^k) = \frac{1}{7}\left(P(1)+C\sum_{1 \leqslant k \leqslant 6} \omega^{4k}\right) = \frac{1}{7}\left(2^{2100}-C\right)\]A small computation results in $C = 2$, therefore the required count is $\frac{2^{2100}-2}{7}$.