If $m$ and $n$ are consecutive positive integers and $n^{2}-m^{2}>20$, then the minimum possible value of $n^{2}+m^{2}$ is
(A) 29
(B) 181
(C) 265
(D) 23
(E) 221
Since $m$ and $n$ are consecutive positive integers with $n^{2}-m^{2}>20$, then $n$ is greater than $m$. Therefore, we can write $n=m+1$.

Since $n^{2}-m^{2}>20$, then $(m+1)^{2}-m^{2}>20$ or $m^{2}+2 m+1-m^{2}>20$ or $2 m>19$ or $m>\frac{19}{2}$. Since $m$ is a positive integer, then $m \geq 10$.

Thus, we want to find the minimum value of $n^{2}+m^{2}=(m+1)^{2}+m^{2}=2 m^{2}+2 m+1$ when $m \geq 10$.

This minimum will occur when $m=10$ (since $2 m^{2}+2 m+1$ increases with $m$ when $m$ is a positive integer).

Therefore, the minimum possible value is $2\left(10^{2}\right)+2(10)+1=221$.

ANSWER: (E)