Let the first digit of the number be $a$ and the number formed by the remaining digits be $k$. By the conditions, $1000a + k = 10k + a + 2016$, whence $111a - k = 224$. Hence $a \ge 3$, implying the solutions $a = 3, k = 109$; $a = 4, k = 220$; $a = 5, k = 331$; $a = 6, k = 442$; $a = 7, k = 553$; $a = 8, k = 664$; $a = 9, k = 775$. The corresponding four-digit numbers satisfying the conditions of the problem are $3109$, $4220$, $5331$, $6442$, $7553$, $8664$, and $9775$.