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Written solution:

The polynomial changes sign at each square $1^2,2^2,\ldots,100^2$, and its leading coefficient is positive. Thus $P(n)\le0$ for integers in the intervals $[1^2,2^2]$, $[3^2,4^2]$, $\ldots$, $[99^2,100^2]$.

For odd $k$, the interval $[k^2,(k+1)^2]$ contains $(k+1)^2-k^2+1=2k+2$ integers. Summing over odd $k=1,3,\ldots,99$ gives $2(1+3+\cdots+99)+2\cdot50=5000+100=5100$. Thus, E is the correct answer.